11.5 校内模拟赛解题报告
作者：互联网
T1
将字符串按长度排序，然后对字符串进行压缩，从 \(a\) 到 \(z\) 排，排满了之后从 \(aa\) 到 \(zz\) 排。
/*
Date:
Source:
Knowledge:
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <algorithm>
#define orz cout << "AK IOI" << "\n"
using namespace std;
const int maxn = 1010;
int read()
{
int x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
while(ch <= '9' && ch >= '0') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
return x * f;
}
void print(int X)
{
if(X < 0) X = ~(X - 1), putchar('-');
if(X > 9) print(X / 10);
putchar(X % 10 ^ '0');
}
int Max(int a, int b){
return a > b ? a : b;
}
int Min(int a, int b){
return a < b ? a : b;
}
int n, num[maxn];
string s[maxn];
map<string, string> mp;
map<int, string> pos;
bool cmp(string a, string b)
{
int len1 = a.length(), len2 = b.length();
return len1 < len2;
}
char tmp[maxn];
void get_ans(int now)
{
int js1 = 0, pos = now, js = 0;
while(now) num[++js] = now % 26, now /= 26;
for(int i = js; i >= 1; i--) tmp[js1++] = 'a' + num[i];
mp[s[pos]] = tmp;
}
int main()
{
n = read();
for(int i = 1; i <= n; i++)
{
cin >> s[i];
pos[i] = s[i];
}
sort(s + 1, s + n + 1, cmp);
int tot = unique(s + 1, s + n + 1) - s - 1;
for(int i = 1; i <= tot; i++) get_ans(i);
for(int i = 1; i <= n; i++) cout << mp[pos[i]], puts(" ");
fclose(stdin);
fclose(stdout);
return 0;
}
T2
用并查集判是否有多余的边，再判断每个点的初度为 \(1\)，有 \(90\) 分。说实话不知道它为啥不对，跟 AC 代码拍了很久也没拍出啥来。
n = read();
for(int i = 1; i <= n; i++) fa[i] = i;
for(int i = 1; i <= n; i++)
{
int u = read(), v = read(), fu = find(u), fv = find(v);
if(in[v] != 0) {printf("%d", i); return 0;}
in[v]++;
if(fu == fv) {printf("%d", i); return 0;}
else fa[fv] = fu;
}
正解：
分类进行讨论。
无环：
找到入度 > 1 的点，把连向它的边编号最大的一个边删掉。
有环：
没有入度为 0 的点，就从环上找一个编号最大的边的删掉，
如果有入度为 0 的点，就找出环上入度为 2 的点，并把连向该点的一条在环上的边删掉。
找环的时候用 tarjan。
/*
Date:2021.11.5
Source:模拟赛
Knowledge: tarjan判环
*/
#include <iostream>
#include <cstdio>
#include <map>
#include <stack>
#define orz cout << "AK IOI" << "\n"
using namespace std;
const int maxn = 1e5 + 10;
int read()
{
int x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
while(ch <= '9' && ch >= '0') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
return x * f;
}
void print(int X)
{
if(X < 0) X = ~(X - 1), putchar('-');
if(X > 9) print(X / 10);
putchar(X % 10 ^ '0');
}
int Max(int a, int b){
return a > b ? a : b;
}
int Min(int a, int b){
return a < b ? a : b;
}
int n, in[maxn], val[maxn], from[maxn], to[maxn], dis[maxn], siz[maxn];
struct node{
int u, v, nxt;
}e[maxn << 1];
int js, head[maxn];
void add(int u, int v)
{
e[++js] = (node){u, v, head[u]};
head[u] = js;
}
int tot, dfn[maxn], low[maxn], vis[maxn];
stack <int> s;
map <int, bool> link[maxn];
void tarjan(int u)
{
s.push(u), vis[u] = 1;
low[u] = dfn[u] = ++tot;
for(int i = head[u]; i; i = e[i].nxt)
{
int v = e[i].v;
if(!dfn[v])
{
tarjan(v);
low[u] = min(low[u], low[v]);
}
else if(vis[v]) low[u] = min(low[u], dfn[v]);
}
int k, lst = u;
if(low[u] == dfn[u])
{
do{
k = s.top(), vis[k] = 0;
s.pop();
link[k][lst] = 1;
lst = k;
if(k == u) break;
dis[u] = max(dis[u], dis[k]);
siz[u]++;
}while(1);
}
}
int flag, id, tag;
int main() {
//freopen("remove.in", "r", stdin);
//freopen("remove.out", "w", stdout);
n = read();
for(int i = 1; i <= n; i++)
{
int u = read(), v = read();
val[v] = max(val[v], i);
from[i] = u, to[i] = v;
in[v]++, dis[v] = i;
add(u, v);
siz[i] = 1;
}
for(int i = 1; i <= n; i++) if(!dfn[i]) tarjan(i);
for(int i = 1; i <= n; i++)
{
if(!in[i]) flag = 1;
if(in[i] == 2) id = i;
}
for(int i = 1; i <= n; i++) if(siz[i] == 1) tag++;
if(tag == n) {print(val[id]); return 0;}
if(flag)
{
for(int i = n; i >= 1; i--)
{
if(link[from[i]][to[i]] && to[i] == id)
{
print(i);
return 0;
}
}
}
for(int i = 1; i <= n; i++)
if(siz[i] > 1) {print(dis[i]); return 0;}
return 0;
}
/*
n = read();
for(int i = 1; i <= n; i++) fa[i] = i;
for(int i = 1; i <= n; i++)
{
int u = read(), v = read(), fu = find(u), fv = find(v);
if(in[v] != 0) {printf("%d", i); return 0;}
in[v]++;
if(fu == fv) {printf("%d", i); return 0;}
else fa[fv] = fu;
}
*/
T3
直接 DP， \(f[i][j]\) 表示甲剩的水量为 \(i\), 已剩的水量为 \(j\) 的概率。
对于 \(k = 1\) 的部分分，打个表会发现，当 n 很大，并且 k=1 的时候，输出都是 1 。
有个结论 \(Ans(n,k) = Ans(\frac{n - 1} 2 + 1, 1)\)。（向上取整）
后面的情况都可以转换到 \(k=1\) 的情况。
/*
Date:
Source:
Knowledge:
*/
#include <iostream>
#include <cstdio>
#define orz cout << "AK IOI" << "\n"
using namespace std;
const int maxn = 1e9 + 10;
int read()
{
int x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
while(ch <= '9' && ch >= '0') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
return x * f;
}
void print(int X)
{
if(X < 0) X = ~(X - 1), putchar('-');
if(X > 9) print(X / 10);
putchar(X % 10 ^ '0');
}
int Max(int a, int b){
return a > b ? a : b;
}
int Min(int a, int b){
return a < b ? a : b;
}
int n, K;
double ans, f[1010][1010];
int main()
{
//freopen("tea.in", "r", stdin);
//freopen("tea.out", "w", stdout);
n = read(), K = read();
n = (n - 1) / K + 1, K = 1; // 向上取整
if(n >= 1000) {puts("1.000000"); return 0;}
f[n][n] = 1.0;
for(int i = n; i >= 1; i--)
{
for(int j = n; j >= 1; j--)
{
for(int k = 1; k <= 4; k++)
f[max(0, i - k * K)][max(0, j - (4 - k) * K)] += f[i][j] / 4.0;
}
}
for(int i = 1; i <= n; i++) ans = ans + f[0][i];
ans = ans + f[0][0] / 2.0;
printf("%.6lf\n", ans);
fclose(stdin);
fclose(stdout);
return 0;
}
标签：ch,return,int,11.5,解题,maxn,low,校内,include 来源： https://www.cnblogs.com/yangchengcheng/p/15514982.html