情人节专场 B题 最短路

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.

Sample test(s)

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

在做这一道时，因为涉及到两两点的距离，我是想到了要用floyd来做。但一来，那时没想清怎么去利用计算每删一个点时的两点之间的最短距离。。。唉，现在才知道要从后往前算，这也算是一个验教训吧。

<span style="color:#333333;">#include<stdio.h>#include<string.h>long long map[501][501],save[501];int vis[501],x[501],n;int main(){freopen("w.txt","r",stdin);int i,j,k;while(~scanf("%d",&n)){memset(vis,0,sizeof(vis));for(i=1;i<=n;i++)for(j=1;j<=n;j++){scanf("%lld",&map[i][j]);}int tot=0;memset(save,0,sizeof(save));for(i=1;i<=n;i++){scanf("%d",&x[i]);}for(i=n;i>=1;i--){vis[x[i]]=1;for(k=1;k<=n;k++)for(j=1;j<=n;j++)if(map[k][j]>map[k][x[i]]+map[x[i]][j])map[k][j]=map[k][x[i]]+map[x[i]][j];for(k=1;k<=n;k++)for(j=1;j<=n;j++)if(vis[k]&&vis[j])save[tot]+=map[k][j];tot++;}for(i=tot-1;i>=1;i--)printf("%lld ",save[i]);printf("%lld\n",save[i]);}return 0;}</span>

17 23 404 0