codeforces 295B Greg and Graph

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 numberx_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 thatd(i, v, u) is the shortest path between verticesv and u in the graph that formed before deleting vertexx_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 linea_ij(1 ≤ a_ij ≤ 10⁵, a_ii = 0) represents the weight of the edge that goes from vertex i to vertexj.

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 inC++. It is preferred to use the cin, cout streams of the %I64d specifier.

Sample Input

Input

101

Output

0 

Input

20 54 01 2

Output

9 0 

Input

40 3 1 16 0 400 12 4 0 11 1 1 04 1 2 3

Output

17 23 404 0 

题意：对于一个带权有向图，每次删除一个点，要求计算出删除该点前当前图的所有点间的最短距离之和。

首先，想知道任意两点间最短距离之和应该使用什么算法？

毫无疑问 Floyd！
对于删除顺序，我们可以怎么处理？

可以反过来做。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。

对于每次新增加的点v，我们用v对整张图进行松弛操作：d[s][t] = min(d[s][t],d[s][v]+d[v][t])

对于统计操作，可以暴力枚举。

整个复杂度为O(n^3)

•注意：本题数据范围可能超过int#include <cstdio>

#include <cstdio>#include <cstring>#define maxn 510typedef long long LL;LL d[maxn][maxn];int seq[maxn];LL ans[maxn];inline LL min(LL a,LL b){    return a < b ? a : b;}void update(int s,int n){    for(int i = 1;i <= n;i++)        for(int j = 1;j <= n;j++)            d[i][j] = min(d[i][s] + d[s][j],d[i][j]);}LL query(int s,int n){    LL sum = 0;    for(int i = n,u = seq[i];i >= s;u = seq[--i])        for(int j = n,v = seq[j];j >= s;v = seq[--j])            sum += d[u][v];    return sum;}int main(){    int n;    while(scanf("%d",&n) != EOF){        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",&seq[i]);        for(int i = n;i >= 1;i--){            update(seq[i],n);            ans[i] = query(i,n);        }        for(int i = 1;i <= n;i++)            printf("%I64d ",ans[i]);        printf("\n");    }    return 0;}

那些年，我们写过的三重循环~