Codeforces 295.B Greg and Graph

B. Greg and Graph

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

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 

 题目大意：给定一个有边权的图。再给定一个1~n的排列。 按排列中的顺序依次删除点，问每次删除后，所有点对的最短路的和是多少。

分析：挺经典的一道题.删点变成加点，利用floyd来更新答案.其实并不需要每次新加一个点以后就再来一次floyd。因为floyd算法的第一层枚举k代表经过k这个点，用它来更新最短路.每次加入一个点实际上就是用这个点来更新其它点对的最短路.每次更新n个点的最短路，但是实际上我们只加入了k个点，所以统计前k个点的答案即可.

　　　要将1~n映射到题目给的排列中去.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 510;

int n,d[maxn][maxn],a[maxn];
long long ans[maxn];

int main()
{
    scanf("%d",&n);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            scanf("%d",&d[i][j]);
    for (int i = 1; i <= n; i++)
        scanf("%d",&a[n - i + 1]);
    for (int k = 1; k <= n; k++)
    {
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
            {
                if (d[a[i]][a[j]] > d[a[i]][a[k]] + d[a[k]][a[j]])
                    d[a[i]][a[j]] = d[a[i]][a[k]] + d[a[k]][a[j]];
            }
        for (int i = 1; i <= k; i++)
            for (int j = 1; j <= k; j++)
                ans[k] += d[a[i]][a[j]];
    }
    for (int i = n; i >= 1; i--)
        cout << ans[i] << " ";

    return 0;
}