E - Help Jimmy POJ - 1661 dp
作者：互联网
E - Help Jimmy
这个题目本身不是很难，但是可以更加优化这个写法。
开始是n*n
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <string>
#define inf 0x3f3f3f3f
#define inf64 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn = 1e3 + 10;
int dp[maxn][2];
struct node {
int l, r, h;
node(int x = 0, int y = 0, int h = 0) :l(x), r(y), h(h) {}
}ex[maxn];
bool cmp(node a, node b) {
return a.h > b.h;
}
int main() {
int t;
scanf("%d", &t);
while (t--) {
int n, x, y, maxs, ans = inf, flag = 0;
memset(dp, inf, sizeof(dp));
scanf("%d%d%d%d", &n, &x, &y, &maxs);
for (int i = 1; i <= n; i++) {
int a, b, w;
scanf("%d%d%d", &a, &b, &w);
if (a > b) swap(a, b);
ex[i] = node(a, b, w);
}
sort(ex + 1, ex + 1 + n, cmp);
for (int i = 1; i <= n; i++) {
if (ex[i].l <= x && ex[i].r >= x && ex[i].h <= y) {
flag = 1;
dp[i][0] = y - ex[i].h + abs(ex[i].l - x);
dp[i][1] = y - ex[i].h + abs(ex[i].r - x);
// printf("l=%d r=%d\n", ex[i].l, ex[i].r);
// printf("dp[%d]=%d dp[%d]=%d %d\n", i, dp[i][0], i, dp[i][1], ex[i].h);
break;
}
}
if (flag == 0 && y <= maxs) {
printf("%d\n", y);
continue;
}
for (int i = 1; i <= n; i++) {
int flag1 = 0, flag2 = 0;
if (dp[i][0] >= inf || dp[i][1] >= inf) continue;
for (int j = 1; j <= n; j++) {
if (i == j) continue;
if (ex[j].h > ex[i].h) continue;
int h = ex[i].h - ex[j].h;
if (h > maxs) continue;
if (ex[i].l <= ex[j].r&&ex[i].l >= ex[j].l&&flag1 == 0) {
flag1 = 1;
dp[j][0] = min(dp[j][0], dp[i][0] + abs(ex[i].l - ex[j].l) + h);
dp[j][1] = min(dp[j][1], dp[i][0] + abs(ex[i].l - ex[j].r) + h);
// printf("l=%d l=%d h=%d\n", ex[i].l, ex[j].l, h);
// printf("ww dp[%d][0]=%d dp[%d][1]=%d %d\n", j, dp[j][0], j, dp[j][1], ex[j].h);
}
if (ex[i].r <= ex[j].r&&ex[i].r >= ex[j].l&&flag2 == 0) {
flag2 = 1;
dp[j][0] = min(dp[j][0], dp[i][1] + abs(ex[i].r - ex[j].l) + h);
dp[j][1] = min(dp[j][1], dp[i][1] + abs(ex[i].r - ex[j].r) + h);
// printf("zz dp[%d][0]=%d dp[%d][1]=%d %d\n", j, dp[j][0], j, dp[j][1], ex[j].h);
}
}
if (flag1 == 0 && ex[i].h <= maxs) ans = min(ans, dp[i][0] + ex[i].h);
if (flag2 == 0 && ex[i].h <= maxs) ans = min(ans, dp[i][1] + ex[i].h);
}
printf("%d\n", ans);
}
return 0;
}
不把起点和终点当成平台的n*n
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <string>
#define inf 0x3f3f3f3f
#define inf64 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn = 1e3 + 10;
int dp[maxn][2];
struct node {
int l, r, h;
node(int x = 0, int y = 0, int h = 0) :l(x), r(y), h(h) {}
}ex[maxn];
bool cmp(node a,node b)
{
return a.h > b.h;
}
int main() {
int t;
scanf("%d", &t);
while (t--) {
int n, x, y, maxs, ans = inf;
memset(dp, inf, sizeof(dp));
scanf("%d%d%d%d", &n, &x, &y, &maxs);
for (int i = 1; i <= n; i++) {
int a, b, w;
scanf("%d%d%d", &a, &b, &w);
if (w >= y) continue;
if (a > b) swap(a, b);
ex[i] = node(a, b, w);
}
ex[++n] = node(x, x, y);
ex[++n] = node(-inf, inf, 0);
sort(ex + 1, ex + 1 + n, cmp);
dp[1][0] = 0, dp[1][1] = 0;
for (int i = 1; i <= n; i++) {
int flag1 = 0, flag2 = 0;
if (dp[i][0] >= inf || dp[i][1] >= inf) continue;
for (int j = i+1; j <= n; j++) {
if (ex[j].h >= ex[i].h) continue;
int h = ex[i].h - ex[j].h;
if (h > maxs) continue;
if (ex[i].l <= ex[j].r&&ex[i].l >= ex[j].l&&flag1 == 0) {
flag1 = 1;
int val1 = abs(ex[i].l - ex[j].l);
int val2 = abs(ex[i].l - ex[j].r);
if (i == n || j == n) val1 = val2 = 0;
dp[j][0] = min(dp[j][0], dp[i][0] + val1 + h);
dp[j][1] = min(dp[j][1], dp[i][0] + val2 + h);
// printf("l=%d l=%d h=%d\n", ex[i].l, ex[j].l, h);
// printf("ww dp[%d][0]=%d dp[%d][1]=%d %d\n", j, dp[j][0], j, dp[j][1], ex[j].h);
}
if (ex[i].r <= ex[j].r&&ex[i].r >= ex[j].l&&flag2 == 0) {
flag2 = 1;
int val1 = abs(ex[i].r - ex[j].l);
int val2 = abs(ex[i].r - ex[j].r);
if (i == n || j == n) val1 = val2 = 0;
dp[j][0] = min(dp[j][0], dp[i][1] + val1 + h);
dp[j][1] = min(dp[j][1], dp[i][1] + val2 + h);
// printf("zz dp[%d][0]=%d dp[%d][1]=%d %d\n", j, dp[j][0], j, dp[j][1], ex[j].h);
}
}
}
printf("%d\n", min(dp[n][0],dp[n][1]));
}
return 0;
}
把起点和终点当成平台的n*n
然后这个题目可以用线段树优化成 n*logn+n 的复杂度。
这个是因为每一个平台会落到的平台是固定的，所以可以用线段树预处理出来每一个平台的下一个平台。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <vector>
#include <string>
#include <algorithm>
#include <iostream>
#include <map>
#define inf 0x3f3f3f3f
#define inf64 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
const int M = 2e4 + 10;
const int maxn = 2e3 + 10;
struct node
{
int id, l, r, lx, rx, y;
node(int id=0,int l=0,int r=0,int lx=0,int rx=0,int y=0):id(id),l(l),r(r),lx(lx),rx(rx),y(y){}
}ex[maxn];
int p[maxn];
ll dp[maxn][2];
bool cmp(int a,int b)
{
return ex[a].y < ex[b].y;
}
bool cmp1(int a, int b) {
return ex[a].y > ex[b].y;
}
int sum[M * 8], lazy[M * 8];
void build(int id,int l,int r)
{
sum[id] = lazy[id] = 0;
if (l == r) return;
int mid = (l + r) >> 1;
build(id << 1, l, mid);
build(id << 1 | 1, mid + 1, r);
}
void push_down(int id)
{
if (lazy[id] == 0) return;
sum[id << 1] = lazy[id];
sum[id << 1 | 1] = lazy[id];
lazy[id << 1] = lazy[id << 1 | 1] = lazy[id];
lazy[id] = 0;
}
void update(int id,int l,int r,int x,int y,int val)
{
// printf("id=%d l=%d r=%d x=%d y=%d\n", id, l, r, x, y);
if(x<=l&&y>=r)
{
lazy[id] = val;
sum[id] = val;
return;
}
push_down(id);
int mid = (l + r) >> 1;
if (x <= mid) update(id << 1, l, mid, x, y, val);
if (y > mid) update(id << 1 | 1, mid + 1, r, x, y, val);
}
int query(int id,int l,int r,int pos)
{
if (l == r) return sum[id];
int mid = (l + r) >> 1;
push_down(id);
if (pos <= mid) return query(id << 1, l, mid, pos);
return query(id << 1 | 1, mid + 1, r, pos);
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
memset(dp, inf, sizeof(dp));
int N, X, Y, H;
scanf("%d%d%d%d", &N, &X, &Y, &H);
X += M;
for(int i=1;i<=N;i++)
{
p[i] = i;
int l, r, h;
scanf("%d%d%d", &l, &r, &h);
l += M, r += M;
ex[i] = node(i, l, r, 0, 0, h);
}
N++;
p[N] = N;
ex[N] = node(0, -inf, inf, 0, 0, 0);
build(1, 1, M * 2);
sort(p + 1, p + 1 + N, cmp);
for(int i=1;i<=N;i++)
{
// printf("i=%d p=%d\n", i, p[i]);
int l = ex[p[i]].l, r = ex[p[i]].r, id = ex[p[i]].id;
ex[p[i]].lx = query(1, 1, M*2, l);
ex[p[i]].rx = query(1, 1, M*2, r);
// printf("l=%d r=%d\n", l, r);
// printf("ex[%d] lx=%d rx=%d\n", p[i], ex[p[i]].lx, ex[p[i]].rx);
update(1, 1, M*2, l, r, id);
}
int num = query(1, 1, M*2, X);
if(num==0)
{
printf("%d\n", Y);
continue;
}
dp[num][0] = Y - ex[num].y + abs(X - ex[num].l);
dp[num][1] = Y - ex[num].y + abs(X - ex[num].r);
// printf("num=%d\n", num);
// printf("dp[%d][0]=%lld dp[%d][1]=%lld\n\n", num, dp[num][0], num, dp[num][1]);
sort(p + 1, p + 1 + N, cmp1);
for(int i=1;i<=N;i++)
{
int id = ex[p[i]].id;
// printf("p[%d]=%d id=%d\n", i, p[i], id);
if (dp[id][0] >= inf) continue;
int lx = ex[id].lx;
int rx = ex[id].rx;
if(ex[id].y-ex[lx].y<=H)
{
int val = abs(ex[id].l - ex[lx].l);
if (lx == 0) val = 0;
dp[lx][0] = min(dp[lx][0], dp[id][0] + ex[id].y - ex[lx].y + val);
val = abs(ex[id].l - ex[lx].r);
if (lx == 0) val = 0;
dp[lx][1] = min(dp[lx][1], dp[id][0] + ex[id].y - ex[lx].y + val);
}
if(ex[id].y-ex[rx].y<=H)
{
int val = abs(ex[id].r - ex[rx].l);
if (rx == 0) val = 0;
dp[rx][0] = min(dp[rx][0], dp[id][1] + ex[id].y - ex[rx].y + val);
val = abs(ex[id].r - ex[rx].r);
if (rx == 0) val = 0;
dp[rx][1] = min(dp[rx][1], dp[id][1] + ex[id].y - ex[rx].y + val);
}
// printf("dp[%d][0]=%lld dp[%d][1]=%lld\n", lx, dp[lx][0], lx, dp[lx][1]);
// printf("dp[%d][0]=%lld dp[%d][1]=%lld\n\n", rx, dp[rx][0], rx, dp[rx][1]);
}
printf("%lld\n", min(dp[0][0], dp[0][1]));
}
return 0;
}
线段树 n*logn+n
还应该可以直接把dp放到线段树里面，就只有n*logn的复杂度。
这个暂时不想写了。
标签：Jimmy,int,POJ,ex,inf,include,id,dp 来源： https://www.cnblogs.com/EchoZQN/p/11385909.html