Problem B Codeforces 295B 最短路(floyd)

標籤：des   style   blog   http   color   io   for   cti   

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 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 ≤ aij ≤ 10^{5, aii = 0)represents the weight of the edge that goes from vertex i to vertex j.}}

The next line contains n distinct integers: x_{1, 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];//存放最短距離 6 int seq[maxn];//刪除的順序 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求最短距離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){//反過來做。這樣就相當於每次添加一個點到圖中，然後詢問當前圖的所有點間最短距離之和。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 }