CF--FLOYD的灵活应用

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 x_i 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 x_i, 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 a_ij (1 ≤ a_ij ≤ 10⁵, a_ii = 0) represents the weight of the edge that goes from vertex i to vertex j.

The next line contains n distinct integers: x₁, x₂, ..., x_n (1 ≤ x_i ≤ 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 test(s)

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 

AC代码：
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <sstream>
#include <vector>
#include <stack>
#include <map>
#include <queue>
#include <set>
#include <iostream>
#include <stdlib.h>
#include <cctype>
#define rep1(i,s,e) for(int i = (s);i<(e);++i)
#define rep2(i,s,e) for(int i = (s);i<=(e);++i)
#define maxn 600
#define INF 1000000007
#define ls (root<<1)
#define rs (root<<1|1)
#define M  L+(R-L)/2
using namespace std;

int n,m;
long long  d[maxn][maxn];
long long sum[maxn];
int del[maxn];
/*
思路：在floyd算法中，三重循环的最外面一重对k的循环的意思
就是增加节点k后更新一遍最短路（更新的最短路都经过节点k）。
那么就可以把最外面一重循环换成v[N,N-1,....,1]，
同样做floyd即可

按照题意是顺序删除反向看就是逆向插入!!!!!!!!
*/
int main()
{
   // freopen("in.txt","r",stdin);
    while(scanf("%d",&n)==1)
    {
        memset(sum,0,sizeof(sum));
        memset(del,0,sizeof(del));
        for(int i = 1;i<=n;++i)
         for(int j = 1;j<=n;++j)
         {
             d[i][j] = INF;
         }

        for(int i = 1;i<=n;++i)
         for(int j = 1;j<=n;++j)
         {
             scanf("%I64d",&d[i][j]);
         }

    for(int i = 1;i<=n;++i)
    scanf("%d",&del[i]);

         //floyd的算法的最外一层是表示把n个点一次插入到图中更新最短路
      int temp  = n;
      for(int k = n;k>=1;--k) //floyd
      {
        for(int i = 1;i<=n;++i)
          for(int j = 1;j<=n;++j)
            d[i][j] = min(d[i][j],d[i][del[k]]+d[del[k]][j]);


        for(int i = n;i>=k;--i)
          for(int j = n;j>=k;--j)
             sum[k]+= d[del[i]][del[j]];
      }

      for(int i = 1;i<=n;++i)
      {
        if(i!= 1)
        printf(" %I64d",sum[i]);
        else
        printf("%I64d",sum[i]);
      }
      printf("\n");

    }
    return 0;
}