Codeforces 296D/295B(Greg and Graph)
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题意:一个有向图,给出每两点间的距离,然后再给出一个删点的序列X,然后求出在删掉Xi前任意两点间的最短路的总和,然后把Xi这个顶点删去,连同和它相连的边;
思路:一道floyd的好题,这题可以倒着做,首先一开始是一个空图,然后往里面加删点序列X里面的点,顺序是删点序列X的逆序,然后每次求出加完点后最短路总和即可。
#include <cstdio>#include <iostream>#include <algorithm>#define LL long longusing namespace std;const int INF = 1234567890;const int N = 505;LL res[N], del[N], dis[N][N], vis[N];int main(){int i, j, k, n;cin >> n;for(i = 1;i <= n;i++)for(j = 1;j <= n;j++)dis[i][j] = INF;for(i = 1;i <= n;i++)for(j = 1;j <= n;j++)cin >> dis[i][j];for(i = 1;i <= n;i++)cin >> del[i];for(k = n;k >= 1;k--){LL sum = 0;vis[del[k]] = 1;for(i = 1;i <= n;i++)for(j = 1;j <= n;j++)dis[i][j] = min(dis[i][j], dis[i][del[k]] + dis[del[k]][j]);for(i = 1;i <= n;i++)for(j = 1;j <= n;j++)if(dis[i][j] != INF && vis[i] && vis[j])sum += dis[i][j];res[k] = sum;}for(i = 1;i <= n;i++)cout << res[i] << " ";cout << endl;return 0;}
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