HDU 1596
Floyd 变形
根据题目中的定义改写新的最短路更新条件
#include <bits/stdc++.h>
using namespace std;
#define endl '
'
typedef long long LL;
const int N = 1e3 + 10;
const double eps = 1e-6;
int n,a,b,m;
double g[N][N],dist[N][N];
void floyd() {
for(int k = 1;k <= n; ++k)
for(int i = 1;i <= n; ++i)
for(int j = 1;j <= n; ++j)
g[i][j] = max(g[i][j],g[i][k] * g[k][j]);
}
int main() {
//ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
while(cin >> n) {
for(int i = 1;i <= n; ++i)
for(int j = 1;j <= n; ++j)
scanf("%lf",&g[i][j]);
floyd();
cin >> m;
while(m --) {
cin >> a >> b;
if(g[a][b] < eps) printf("What a pity!
");
else printf("%.3lf
",g[a][b]);
}
}
return 0;
}