题目链接

思路

floyd求一下传递闭包，然后统计每个点可以到达的点数。
会tle，用bitset优化一下。将floyd的最后一层枚举变成bitset。

代码

/*
* @Author: wxyww
* @Date:   2019-01-23 15:08:40
* @Last Modified time: 2019-01-23 15:22:52
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<bitset>
using namespace std;
typedef long long ll;
const int N = 2010;
bitset<N>f[N];
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*10+c-'0';
      c=getchar();
   }
   return x*f;
}
char s[N];
int main() {
   int n = read();
   for(int i = 1;i <= n;++i) {
      scanf("%s",s + 1);
      for(int j = 1;j <= n;++j)        
         f[i][j] = s[j] - '0';
      f[i][i] = 1;
   }
   for(int k = 1;k <= n;++k) 
      for(int i = 1;i <= n;++i) 
         if(f[i][k]) f[i] |= f[k];
   ll ans = 0;
   for(int i = 1;i <= n;++i)
      ans += f[i].count();
   cout<<ans;
   return 0;
}