首页 > 代码库 > Problem B Codeforces 295B 最短路(floyd)

Problem B Codeforces 295B 最短路(floyd)

2024-07-15 15:51:57   222人阅读

Description

Greg has a weighed directed graph, consisting of n vertices. In this graph any pair of distinct vertices has an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game:

  • The game consists of n steps.
  • On the i-th step Greg removes vertex number xi from the graph. As Greg removes a vertex, he also removes all the edges that go in and out of this vertex.
  • Before executing each step, Greg wants to know the sum of lengths of the shortest paths between all pairs of the remaining vertices. The shortest path can go through any remaining vertex. In other words, if we assume that d(i, v, u) is the shortest path between vertices v and u in the graph that formed before deleting vertex xi, then Greg wants to know the value of the following sum: .

Help Greg, print the value of the required sum before each step.

Input

The first line contains integer n(1 ≤ n ≤ 500) — the number of vertices in the graph.

Next n lines contain n integers each — the graph adjacency matrix: the j-th number in the i-th line aij(1 ≤ aij ≤ 105, aii = 0)represents the weight of the edge that goes from vertex i to vertex j.

The next line contains n distinct integers: x1, x2, ..., xn(1 ≤ xi ≤ n) — the vertices that Greg deletes.

Output

Print n integers — the i-th number equals the required sum before the i-th step.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64dspecifier.

Sample Input

Input
1
0
1
Output
0
Input
2
0 5
4 0
1 2
Output
9 0
Input
4
0 3 1 1
6 0 400 1
2 4 0 1
1 1 1 0
4 1 2 3
Output
17 23 404 0 
 1 #include <cstdio> 2 #include <cstring> 3 #define maxn 510 4 typedef long long LL; 5 LL d[maxn][maxn];//存放最短距离 6 int seq[maxn];//删除的顺序 7 LL ans[maxn];//答案 8  9 inline LL min(LL a,LL b){10     return a < b ? a : b;11 }12 13 void update(int s,int n){//floyd求最短距离14     for(int i = 1;i <= n;i++)15         for(int j = 1;j <= n;j++)16             d[i][j] = min(d[i][s] + d[s][j],d[i][j]);17 }18 19 LL query(int s,int n){//反过来做。这样就相当于每次添加一个点到图中，然后询问当前图的所有点间最短距离之和。20     LL sum = 0;21     for(int i = n,u = seq[i];i >= s;u = seq[--i])//傻傻看不懂22         for(int j = n,v = seq[j];j >= s;v = seq[--j])23             sum += d[u][v];24     return sum;25 }26 27 int main()28 {29     int n;30     while(scanf("%d",&n) != EOF){31         for(int i = 1;i <= n;i++)32             for(int j = 1;j <= n;j++)33                 scanf("%I64d",&d[i][j]);34         for(int i = 1;i <= n;i++)35             scanf("%d",&seq[i]);36         for(int i = n;i >= 1;i--){37             update(seq[i],n);38             ans[i] = query(i,n);39         }40         for(int i = 1;i <= n;i++)41             printf("%I64d ",ans[i]);42         printf("\n");43     }44     return 0;45 }

 


联系
我们