《搜索专题练习-题解》
作者：互联网
A:
bfs搜索即可，每次都对当前的数字进行所有的变换，然后记录一下状态有没有出现过，第一次遇到那个数字一定是时间最小的。
这里我的写法多次一举写了个小顶堆的bfs。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 1e5 + 5;
const int M = 5e5 + 5;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18 + 5
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
inline int read(){
int x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
return x*f;
}
}
using namespace FASTIO;
int dp[10][10][10][10];
int cal1(int x) {
if(x == 9) return 1;
else return x + 1;
}
int cal2(int x) {
if(x == 1) return 9;
else return x - 1;
}
struct Node{
int x1,x2,x3,x4,sum;
bool operator < (const Node a) const {
return sum > a.sum;
}
};
void bfs(int x1,int x2,int x3,int x4) {
priority_queue<Node> Q;
dp[x1][x2][x3][x4] = 0;
Q.push(Node{x1,x2,x3,x4,0});
while(!Q.empty()) {
Node q = Q.top();
int px1 = q.x1,px2 = q.x2,px3 = q.x3,px4 = q.x4;
x1 = q.x1,x2 = q.x2,x3 = q.x3,x4 = q.x4;
int sum = q.sum;
Q.pop();
px1 = cal1(x1);
if(sum + 1 < dp[px1][px2][px3][px4]) {
dp[px1][px2][px3][px4] = sum + 1;
Q.push(Node{px1,px2,px3,px4,sum + 1});
}
px1 = x1;
px1 = cal2(x1);
if(sum + 1 < dp[px1][px2][px3][px4]) {
dp[px1][px2][px3][px4] = sum + 1;
Q.push(Node{px1,px2,px3,px4,sum + 1});
}
px1 = x1;
px2 = cal1(x2);
if(sum + 1 < dp[px1][px2][px3][px4]) {
dp[px1][px2][px3][px4] = sum + 1;
Q.push(Node{px1,px2,px3,px4,sum + 1});
}
px2 = x2;
px2 = cal2(x2);
if(sum + 1 < dp[px1][px2][px3][px4]) {
dp[px1][px2][px3][px4] = sum + 1;
Q.push(Node{px1,px2,px3,px4,sum + 1});
}
px2 = x2;
px3 = cal1(x3);
if(sum + 1 < dp[px1][px2][px3][px4]) {
dp[px1][px2][px3][px4] = sum + 1;
Q.push(Node{px1,px2,px3,px4,sum + 1});
}
px3 = x3;
px3 = cal2(x3);
if(sum + 1 < dp[px1][px2][px3][px4]) {
dp[px1][px2][px3][px4] = sum + 1;
Q.push(Node{px1,px2,px3,px4,sum + 1});
}
px3 = x3;
px4 = cal1(x4);
if(sum + 1 < dp[px1][px2][px3][px4]) {
dp[px1][px2][px3][px4] = sum + 1;
Q.push(Node{px1,px2,px3,px4,sum + 1});
}
px4 = x4;
px4 = cal2(x4);
if(sum + 1 < dp[px1][px2][px3][px4]) {
dp[px1][px2][px3][px4] = sum + 1;
Q.push(Node{px1,px2,px3,px4,sum + 1});
}
px4 = x4;
px1 = x2,px2 = x1;
if(sum + 1 < dp[px1][px2][px3][px4]) {
dp[px1][px2][px3][px4] = sum + 1;
Q.push(Node{px1,px2,px3,px4,sum + 1});
}
px1 = x1,px2 = x2;
px2 = x3,px3 = x2;
if(sum + 1 < dp[px1][px2][px3][px4]) {
dp[px1][px2][px3][px4] = sum + 1;
Q.push(Node{px1,px2,px3,px4,sum + 1});
}
px2 = x2,px3 = x3;
px3 = x4,px4 = x3;
if(sum + 1 < dp[px1][px2][px3][px4]) {
dp[px1][px2][px3][px4] = sum + 1;
Q.push(Node{px1,px2,px3,px4,sum + 1});
}
px3 = x3,px4 = x4;
}
}
int main()
{
int ca = read();
while(ca--) {
memset(dp,0x3f3f3f,sizeof(dp));
string s1,s2;
cin >> s1 >> s2;
bfs(s1[0] - '0',s1[1] - '0',s1[2] - '0',s1[3] - '0');
printf("%d\n",dp[s2[0] - '0'][s2[1] - '0'][s2[2] - '0'][s2[3] - '0']);
}
return 0;
}
View Code
B：
依旧是bfs搜索，可以发现这里只有人的位置和箱子的位置在变，那么就用vis[i][j][k][m]表示当箱子的位置在i,j，人的位置在k,m时，有没有出现过这个状态。
然后注意推的时候看看人能不能走到推的那个位置，这里用dfs来搜能不能走到。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 1e5 + 5;
const int M = 5e5 + 5;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18 + 5
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
inline int read(){
int x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
return x*f;
}
}
using namespace FASTIO;
int a[10][10],n,m,b[4][2] = {1,0,-1,0,0,1,0,-1};
bool vis[10][10][10][10],used[10][10];
struct Node {
int x,y,dis,prex,prey;
bool operator < (const Node a) const {
return dis > a.dis;
}
};
bool dfs(int x,int y,int zx,int zy,int ex,int ey) {
used[x][y] = 1;
if(x == ex && y == ey) return true;
int ans = 0;
for(int i = 0;i < 4;++i) {
int px = x + b[i][0];
int py = y + b[i][1];
if(px >= 1 && px <= n && py >= 1 && py <= m && !used[px][py] && a[px][py] != 1 && (px != zx || py != zy)) {
ans |= dfs(px,py,zx,zy,ex,ey);
}
}
return ans;
}
int bfs(int sx,int sy,int ex,int ey,int posx,int posy) {
priority_queue<Node> Q;
Q.push(Node{sx,sy,0,posx,posy});
while(!Q.empty()) {
Node q = Q.top();
Q.pop();
if(q.x == ex && q.y == ey) return q.dis;
for(int i = 0;i < 4;++i) {
int px = q.x + b[i][0];
int py = q.y + b[i][1];
int pxx = q.x - b[i][0];
int pyy = q.y - b[i][1];
if(px >= 1 && px <= n && py >= 1 && py <= m && a[px][py] != 1 && pxx >= 1 && pxx <= n && pyy >= 1 && pyy <= m && a[pxx][pyy] != 1 && !vis[px][py][pxx][pyy]) {
memset(used,0,sizeof(used));
// printf("(%d,%d) peo(%d,%d) to(%d,%d)\n",q.x,q.y,q.prex,q.prey,pxx,pyy);
if(dfs(q.prex,q.prey,q.x,q.y,pxx,pyy)) {
// printf("YES\n");
vis[px][py][pxx][pyy] = 1;
Q.push(Node{px,py,q.dis + 1,q.x,q.y});
}
else {
// printf("no\n");
}
}
}
}
return -1;
}
int main()
{
int ca;ca = read();
while(ca--) {
n = read(),m = read();
memset(vis,0,sizeof(vis));
int sx,sy,ex,ey,posx,posy;
for(int i = 1;i <= n;++i)
for(int j = 1;j <= m;++j) a[i][j] = read();
for(int i = 1;i <= n;++i) {
for(int j = 1;j <= m;++j) {
if(a[i][j] == 2) sx = i,sy = j;
if(a[i][j] == 3) ex = i,ey = j;
if(a[i][j] == 4) posx = i,posy = j;
}
}
printf("%d\n",bfs(sx,sy,ex,ey,posx,posy));
}
return 0;
}
/*2
5 5
0 3 0 0 0
1 0 1 4 0
0 0 1 0 0
1 1 2 0 0
0 0 0 0 0
*/
View Code
C：
双向bfs搜索，因为这里要保证8步内到，且起点和终点状态都确定，那么我们就可以从起点和终点同时开始bfs，注意我们让两边的时间限制都为4，这样当一边的状态
在另一个队列出现，且时间没超过4的情况，就说明能走到，注意这里int开不出空间，所以就用char来开。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 1e5 + 5;
const int M = 5e5 + 5;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18 + 5
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
inline int read(){
int x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
return x*f;
}
}
using namespace FASTIO;
struct Node {
int x,y;
bool operator == (const Node a) const {
if(a.x == x && a.y == y) return true;
else return false;
}
bool operator != (const Node a) const {
if(a.x != x || a.y != y) return true;
else return false;
}
};
Node a[5],b[5];
char vis[9][9][9][9][9][9][9][9];
int dir[8][2] = {1,0,-1,0,0,1,0,-1,2,0,-2,0,0,2,0,-2};
struct PP {
Node p1,p2,p3,p4;
int dis;
PP(Node a,Node b,Node c,Node d,int x) {
p1 = a,p2 = b,p3 = c,p4 = d,dis = x;
}
};
int check1(int x,int y,PP a) {
if(x >= 0 && x < 8 && y >= 0 && y < 8 && a.dis <= 4) {
if(vis[a.p1.x][a.p1.y][a.p2.x][a.p2.y][a.p3.x][a.p3.y][a.p4.x][a.p4.y] == '1') return 0;
else if(vis[a.p1.x][a.p1.y][a.p2.x][a.p2.y][a.p3.x][a.p3.y][a.p4.x][a.p4.y] == '2') return 1;
else {
vis[a.p1.x][a.p1.y][a.p2.x][a.p2.y][a.p3.x][a.p3.y][a.p4.x][a.p4.y] = '1';
return 2;
}
}
return 0;
}
int check2(int x,int y,PP a) {
if(x >= 0 && x < 8 && y >= 0 && y < 8 && a.dis <= 4) {
if(vis[a.p1.x][a.p1.y][a.p2.x][a.p2.y][a.p3.x][a.p3.y][a.p4.x][a.p4.y] == '2') return 0;
else if(vis[a.p1.x][a.p1.y][a.p2.x][a.p2.y][a.p3.x][a.p3.y][a.p4.x][a.p4.y] == '1') return 1;
else {
vis[a.p1.x][a.p1.y][a.p2.x][a.p2.y][a.p3.x][a.p3.y][a.p4.x][a.p4.y] = '2';
return 2;
}
}
return 0;
}
bool bfs() {
queue<PP> Q1,Q2;
vis[a[1].x][a[1].y][a[2].x][a[2].y][a[3].x][a[3].y][a[4].x][a[4].y] = '1';
vis[b[1].x][b[1].y][b[2].x][b[2].y][b[3].x][b[3].y][b[4].x][b[4].y] = '2';
Q1.push(PP{Node{a[1].x,a[1].y} , Node{a[2].x,a[2].y} , Node{a[3].x,a[3].y} , Node{a[4].x,a[4].y} , 0});
Q2.push(PP{Node{b[1].x,b[1].y} , Node{b[2].x,b[2].y} , Node{b[3].x,b[3].y} , Node{b[4].x,b[4].y} , 0});
while(!Q1.empty() || !Q2.empty()) {
if(!Q1.empty()) {
PP q = Q1.front();
Q1.pop();
for(int i = 0;i < 8;++i) {
PP t = q;
int px = q.p1.x + dir[i][0];
int py = q.p1.y + dir[i][1];
t.p1 = Node{px,py};
t.dis = q.dis + 1;
int ma = check1(px,py,t);
if(ma == 1) return true;
if(ma == 2) {
Q1.push(t);
}
}
for(int i = 0;i < 8;++i) {
PP t = q;
int px = q.p2.x + dir[i][0];
int py = q.p2.y + dir[i][1];
t.p2 = Node{px,py};
t.dis = q.dis + 1;
int ma = check1(px,py,t);
if(ma == 1) return true;
if(ma == 2) {
Q1.push(t);
}
}
for(int i = 0;i < 8;++i) {
PP t = q;
int px = q.p3.x + dir[i][0];
int py = q.p3.y + dir[i][1];
t.p3 = Node{px,py};
t.dis = q.dis + 1;
int ma = check1(px,py,t);
if(ma == 1) return true;
if(ma == 2) {
Q1.push(t);
}
}
for(int i = 0;i < 8;++i) {
PP t = q;
int px = q.p4.x + dir[i][0];
int py = q.p4.y + dir[i][1];
t.p4 = Node{px,py};
t.dis = q.dis + 1;
int ma = check1(px,py,t);
if(ma == 1) return true;
if(ma == 2) {
Q1.push(t);
}
}
}
if(!Q2.empty()) {
PP q = Q2.front();
Q2.pop();
for(int i = 0;i < 8;++i) {
PP t = q;
int px = q.p1.x + dir[i][0];
int py = q.p1.y + dir[i][1];
t.p1 = Node{px,py};
t.dis = q.dis + 1;
int ma = check2(px,py,t);
if(ma == 1) return true;
if(ma == 2) {
Q2.push(t);
}
}
for(int i = 0;i < 8;++i) {
PP t = q;
int px = q.p2.x + dir[i][0];
int py = q.p2.y + dir[i][1];
t.p2 = Node{px,py};
t.dis = q.dis + 1;
int ma = check2(px,py,t);
if(ma == 1) return true;
if(ma == 2) {
Q2.push(t);
}
}
for(int i = 0;i < 8;++i) {
PP t = q;
int px = q.p3.x + dir[i][0];
int py = q.p3.y + dir[i][1];
t.p3 = Node{px,py};
t.dis = q.dis + 1;
int ma = check2(px,py,t);
if(ma == 1) return true;
if(ma == 2) {
Q2.push(t);
}
}
for(int i = 0;i < 8;++i) {
PP t = q;
int px = q.p4.x + dir[i][0];
int py = q.p4.y + dir[i][1];
t.p4 = Node{px,py};
t.dis = q.dis + 1;
int ma = check2(px,py,t);
if(ma == 1) return true;
if(ma == 2) {
Q2.push(t);
}
}
}
}
return false;
}
int main()
{
while(~scanf("%d",&a[1].x)) {
memset(vis,0,sizeof(vis));
a[1].x--;
a[1].y = read() - 1;
for(int i = 2;i <= 4;++i) a[i].x = read() - 1,a[i].y = read() - 1;
for(int i = 1;i <= 4;++i) b[i].x = read() - 1,b[i].y = read() - 1;
printf("%s\n",bfs() ? "YES" : "NO");
}
return 0;
}
View Code
D：
这里题目的限制描述有点不太严谨，应该保证无环产生才行。
这样的话就是一棵树，那么题目问的也就是树的直径。
树的直径有一个性质：到某个点距离最远的点肯定是直径上的一个点。
那么我们就可以以1为根找到和1最远的点，肯定就是直径上的一个点。
那么再以这个点为根去找到直径上的另一个点，他们的距离就是直径了。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 4e4 + 5;
const int M = 88888888;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18 + 5
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
inline LL read(){
LL x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
return x*f;
}
}
using namespace FASTIO;
int n,m,dp[N];//0 - xia,1 - shang
struct Node{
int to,dis;
};
vector<Node> G[N];
void dfs(int u,int fa,int d) {
dp[u] = dp[fa] + d;
for(auto v : G[u]) {
if(v.to == fa) continue;
dfs(v.to,u,v.dis);
}
}
int main()
{
n = read(),m = read();
while(m--) {
int x,y,L;x = read(),y = read(),L = read();
char c;cin >> c;
G[x].push_back(Node{y,L});
G[y].push_back(Node{x,L});
}
dfs(1,0,0);
int mx = 0,pos = 0;
for(int i = 1;i <= n;++i) {
if(dp[i] > mx) {
mx = dp[i];
pos = i;
}
}
memset(dp,0,sizeof(dp));
dfs(pos,0,0);
int ans = 0;
for(int i = 1;i <= n;++i) {
ans = max(ans,dp[i]);
}
printf("%d\n",ans);
return 0;
}
View Code
E：
这题主要是卡空间和时间，因为这里说1不能移动。
那么我们把这个环看成以1开头的一个数字，且我们保证状态永远以1作为开头。
dp[x]表示1后面顺时针接数字x的最小步数。
这里1不需要记录所以最大状态就是87654321,空间和时间复杂度都压下来了。
因为我们最终都是要得到12345678，那么我们就可以一开始从12345678倒着搜，预处理出所有状态的最小步数。
预处理也是基本的bfs，要记录一下在中心的点是谁就行了。
因为我们保证了
这样询问的时候就可以O（1）输出。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 1e5 + 5;
const int M = 88888888;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18 + 5
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
inline LL read(){
LL x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
return x*f;
}
}
using namespace FASTIO;
int a[10];
int dp[8765432 + 1];
struct Node{
int x,dis,in;
};
void init() {
dp[2345678] = 1;
queue<Node> Q;
Q.push(Node{2345678,1,0});
while(!Q.empty()) {
Node q = Q.front();
int c = q.x;
Q.pop();
//printf("%d is %d\n",c,q.dis);
int pos0 = 0;
for(int i = 7;i >= 1;--i) {
a[i] = q.x % 10;
q.x /= 10;
if(a[i] == 0) pos0 = i;
}
//if(pos0 == 0) printf("%d is %d\n",c,q.dis);
if(q.in == 0) {
for(int i = 1;i <= 7;++i) {
int ma = 0;
for(int j = 1;j <= 7;++j) {
if(j == i) ma = (ma * 10 + 0);
else ma = ma * 10 + a[j];
}
if(dp[ma] == 0) {
dp[ma] = q.dis + 1;
Q.push(Node{ma,q.dis + 1,a[i]});
}
}
}
else {
int ma = 0;//in
for(int i = 1;i <= 7;++i) {
if(a[i] == 0) ma = (ma * 10 + q.in);
else ma = (ma * 10 + a[i]);
}
if(dp[ma] == 0) {
dp[ma] = q.dis + 1;
Q.push(Node{ma,q.dis + 1,0});
}
if(pos0 >= 2 && pos0 <= 6) {
ma = 0;//left
for(int i = 1;i <= 7;++i) {
if(i == pos0 - 1) ma = (ma * 10 + 0);
else if(i == pos0) ma = (ma * 10 + a[i - 1]);
else ma = ma * 10 + a[i];
}
if(dp[ma] == 0) {
dp[ma] = q.dis + 1;
Q.push(Node{ma,q.dis + 1,q.in});
}
ma = 0;//right
for(int i = 1;i <= 7;++i) {
if(i == pos0 + 1) ma = (ma * 10 + 0);
else if(i == pos0) ma = (ma * 10 + a[i + 1]);
else ma = ma * 10 + a[i];
}
if(dp[ma] == 0) {
dp[ma] = q.dis + 1;
Q.push(Node{ma,q.dis + 1,q.in});
}
}
else if(pos0 == 1){
ma = a[2];
ma = (ma * 10 + 0);
for(int i = 3;i <= 7;++i) {
ma = (ma * 10 + a[i]);
}
if(dp[ma] == 0) {
dp[ma] = q.dis + 1;
Q.push(Node{ma,q.dis + 1,q.in});
}
}
else {
ma = 0;
for(int i = 1;i <= 5;++i) {
ma = (ma * 10 + a[i]);
}
ma = (ma * 10 + 0);
ma = (ma * 10 + a[6]);
if(dp[ma] == 0) {
dp[ma] = q.dis + 1;
Q.push(Node{ma,q.dis + 1,q.in});
}
}
}
}
}
int cal(int x) {
if(x == 8) return 1;
else return x + 1;
}
int main()
{
init();
int ca;ca = read();
while(ca--) {
int pos;
for(int i = 1;i <= 8;++i) a[i] = read();
for(int i = 1;i <= 8;++i) {
if(a[i] == 1) pos = i;
}
int i = cal(pos);
int ma = a[i];
while(cal(i) != pos) {
i = cal(i);
ma = (ma * 10 + a[i]);
}
// dbg(ma);
printf("%d\n",dp[ma] - 1);
}
return 0;
}
View Code
F：
依旧是bfs搜索，vis[i][j][k][m] - 表示当空白位置为i,j，指定棋子为k,m时是否被搜到过。
然后bfs的时候我们取移动空白的位置即可。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 1e5 + 5;
const int M = 5e5 + 5;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18 + 5
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
inline LL read(){
LL x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
return x*f;
}
}
using namespace FASTIO;
int n,m,q,ex,ey,sx,sy,tx,ty,b[4][2] = {1,0,-1,0,0,1,0,-1},a[35][35];
bool vis[35][35][35][35];
struct Node{
int x1,y1,x2,y2,dis;
};
int solve() {
memset(vis,0,sizeof(vis));
queue<Node> Q;
Q.push(Node{ex,ey,sx,sy,0});
vis[ex][ey][sx][sy] = 1;
while(!Q.empty()) {
Node q = Q.front();
Q.pop();
if(q.x2 == tx && q.y2 == ty) return q.dis;
for(int i = 0;i < 4;++i) {
int px = q.x1 + b[i][0];
int py = q.y1 + b[i][1];
if(px >= 1 && px <= n && py >= 1 && py <= m && a[px][py] != 0) {
if(px == q.x2 && py == q.y2) {
if(!vis[q.x2][q.y2][q.x1][q.y1]) {
//printf("(%d,%d) (%d,%d) to (%d,%d) (%d,%d) pos1\n",q.x1,q.y1,q.x2,q.y2,q.x2,q.y2,q.x1,q.y1);
vis[q.x2][q.y2][q.x1][q.y1] = 1;
Q.push(Node{q.x2,q.y2,q.x1,q.y1,q.dis + 1});
}
}
else if(!vis[px][py][q.x2][q.y2]){
//printf("(%d,%d) (%d,%d) to (%d,%d) (%d,%d) pos2\n",q.x1,q.y1,q.x2,q.y2,px,py,q.x2,q.y2);
vis[px][py][q.x2][q.y2] = 1;
Q.push(Node{px,py,q.x2,q.y2,q.dis + 1});
}
}
}
}
return -1;
}
int main()
{
n = read(),m = read(),q = read();
for(int i = 1;i <= n;++i)
for(int j = 1;j <= m;++j) a[i][j] = read();
while(q--) {
ex = read(),ey = read(),sx = read(),sy = read(),tx = read(),ty = read();
printf("%d\n",solve());
}
return 0;
}
/*
3 4 1
0 1 1 1
0 1 1 0
0 1 0 0
1 2 1 4 3 2
*/
View Code
G：
我们把一个空洞看成一个点，能到达的空洞之间建边，然后再从最下面开始bfs即可。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 1e5 + 5;
const int M = 5e5 + 5;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18 + 5
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
inline LL read(){
LL x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
return x*f;
}
}
using namespace FASTIO;
LL x[1005],y[1005],z[1005];
int dis[1005][1005];
int n,h,r;
bool vis[1005];
LL cal(int i,int j) {
return (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]) + (z[i] - z[j]) * (z[i] - z[j]);
}
bool bfs() {
queue<int> Q;
for(int i = 1;i <= n;++i) {
if(z[i] <= r) {
Q.push(i);
vis[i] = 1;
}
}
while(!Q.empty()) {
int u = Q.front();
// printf("%d\n",u);
Q.pop();
if(z[u] + r >= h) return true;
for(int i = 1;i <= n;++i) {
if(!vis[i] && dis[u][i] == 1) {
vis[i] = 1;
Q.push(i);
}
}
}
return false;
}
int main()
{
int ca;ca = read();
while(ca--) {
memset(dis,0,sizeof(dis));
memset(vis,0,sizeof(vis));
n = read(),h = read(),r = read();
for(int i = 1;i <= n;++i) x[i] = read(),y[i] = read(),z[i] = read();
for(int i = 1;i <= n;++i) {
for(int j = 1;j <= n;++j) {
if(i == j) continue;
LL ma = cal(i,j);
if(cal(i,j) <= 4LL * r * r) {
dis[i][j] = dis[j][i] = 1;
}
}
}
printf("%s\n",bfs() ? "Yes" : "No");
}
return 0;
}
View Code
H：
首先，我们需要知道，一个第一行的点能到达的最后一行的所有点一定是连起来的，不然中间的点别的点都无法到达。
那么我们就可以处理出每个点能到达的一条连续的区域，我们可以看成一条线段。那么最后就是一个区间覆盖的问题。
我们对所有线段左端点升序，然后贪心去找最远的右端点覆盖即可。
dfs处理每个点能到达的最左的点和最右的点。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 1e5 + 5;
const int M = 5e5 + 5;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18 + 5
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
inline int read(){
int x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
return x*f;
}
}
using namespace FASTIO;
int n,m,a[505][505],vis[505][505],le[505][505],ri[505][505],b[4][2] = {1,0,-1,0,0,1,0,-1};
struct Node{
int L,r;
bool operator < (const Node a) const {
if(L == a.L) return r > a.r;
else return L < a.L;
}
}p[505];
void dfs(int x,int y) {
if(vis[x][y]) return ;
vis[x][y] = 1;
if(x == n) {
le[x][y] = ri[x][y] = y;
}
for(int i = 0;i < 4;++i) {
int px = x + b[i][0];
int py = y + b[i][1];
if(px >= 1 && px <= n && py >= 1 && py <= m && a[px][py] < a[x][y]) {
if(!vis[px][py]) {
dfs(px,py);
}
le[x][y] = min(le[px][py],le[x][y]);
ri[x][y] = max(ri[px][py],ri[x][y]);
}
}
}
int main()
{
while(~scanf("%d %d",&n,&m)) {
for(int i = 1;i <= n;++i)
for(int j = 1;j <= m;++j) {
a[i][j] = read();
vis[i][j] = ri[i][j] = 0;
le[i][j] = 1000000000;
}
for(int i = 1;i <= m;++i) {
if(vis[1][i]) continue;
dfs(1,i);
}
int cnt = 0;
for(int i = 1;i <= m;++i) {
if(vis[n][i] == 0) cnt++;
}
if(cnt == 0) {
for(int i = 1;i <= m;++i) p[i].L = le[1][i],p[i].r = ri[1][i];
sort(p + 1,p + m + 1);
/* for(int i = 1;i <= m;++i) {
printf("p[%d].L is %d p[%d].r is %d\n",i,p[i].L,i,p[i].r);
}*/
int st = 2,ans = 1,ri = p[1].r;
while(ri < m && st <= m) {
int mx = 0;
while(p[st].L <= ri + 1 && st <= m) {
mx = max(mx,p[st++].r);
}
ri = max(ri,mx);
ans++;
}
printf("1\n%d\n",ans);
}
else {
printf("0\n%d\n",cnt);
}
}
return 0;
}
View Code
标签：Node,专题,int,题解,px1,练习,px4,px3,px2 来源： https://www.cnblogs.com/zwjzwj/p/14736381.html