那些年我们写过的三重循环----CodeForces 295B Greg and Graph 重温Floyd算法

最新推荐文章于 2019-09-21 15:04:35 发布

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#Floyd #dp #Greg and Graph

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本文介绍了一个基于图论的经典算法题目，通过GregandGraph游戏的形式展现Floyd算法的应用。在一个完全加权有向图中，玩家需要移除顶点并计算剩余顶点间最短路径长度之和。

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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: $\text{[math]}$ .

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 thecin, cout streams of the %I64d specifier.

   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 

题目链接：here

非常值得一做的经典题目，让我们再一次想起Floyd算法三重for循环背后的光芒。

#include<cstdio>
#define N 505
#define ll long long int

ll dp[N][N],arr[N],ans[N],n;

ll _min(ll a,ll b) {
    return a<b?a:b;
}

int main()
{
    while(scanf("%d",&n)!=EOF) {
        for(int i=0;i<n;i++) {
            for(int j=0;j<n;j++) scanf("%lld",&dp[i][j]);
        }
        for(int i=0;i<n;i++) scanf("%lld",&arr[n-i-1]);
        for(int i=0;i<n;i++) arr[i]--;
        for(int mid=0;mid<n;mid++) {
            int u=arr[mid];
            ll temp=0;
            for(int e=0;e<n;e++) {
                for(int s=0;s<n;s++) {
                    int a=arr[s];
                    int b=arr[e];
                    dp[a][b]=_min(dp[a][b],dp[a][u]+dp[u][b]);
                    if(s<=mid && e<=mid) temp+=dp[a][b];
                }
            }
            ans[n-mid-1]=temp;
        }
        for(int i=0;i<n;i++) printf("%I64d ",ans[i]);
        printf("\n");
    }
    return 0;
}