Hdu 6026 Deleting Edges【思维+最短路】

Deleting Edges

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 456    Accepted Submission(s): 162

Problem Description

Little Q is crazy about graph theory, and now he creates a game about graphs and trees.
There is a bi-directional graph with n nodes, labeled from 0 to n−1. Every edge has its length, which is a positive integer ranged from 1 to 9.
Now, Little Q wants to delete some edges (or delete nothing) in the graph to get a new graph, which satisfies the following requirements:
(1) The new graph is a tree with n−1 edges.
(2) For every vertice v(0<v<n), the distance between 0 and v on the tree is equal to the length of shortest path from 0 to v in the original graph.
Little Q wonders the number of ways to delete edges to get such a satisfied graph. If there exists an edge between two nodesi and j, while in another graph there isn't such edge, then we regard the two graphs different.
Since the answer may be very large, please print the answer modulo 109+7.

 

Input

The input contains several test cases, no more than 10 test cases.
In each test case, the first line contains an integer n(1≤n≤50), denoting the number of nodes in the graph.
In the following n lines, every line contains a string with n characters. These strings describes the adjacency matrix of the graph. Suppose thej-th number of the i-th line is c(0≤c≤9), if c is a positive integer, there is an edge between i and j with length of c, if c=0, then there isn't any edge between i and j.
The input data ensure that the i-th number of the i-th line is always 0, and the j-th number of the i-th line is always equal to the i-th number of the j-th line.

 

Output

For each test case, print a single line containing a single integer, denoting the answer modulo109+7.

 

Sample Input

2011040123101221013210

 

Sample Output

16

题目大意：

给你N个点，然后给你一个邻接矩阵，让你求一共有多少种删除边的方式，使得构成一个生成树之后，从0到任意点v的距离都是原图的最短路的距离。

思路：

首先我们预处理出从点0到其他各个点的最短路的距离dist【i】.

那么对应如果有dist【i】+w==dist【j】，那么从i到j这条边就可以作为构成生成树的一条边。

那么我们将所有可以未做生成树的边的个数处理出来，那么对应在每个点上边取一条边出来作为答案即可。

那么累成结果就行了。

Ac代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<queue>using namespace std;char a[55][55];int dist[55];int vis[55];int ans[55];int n;void SPFA(){    memset(vis,0,sizeof(vis));    queue<int >s;    for(int i=0;i<n;i++)dist[i]=0x3f3f3f3f;    dist[0]=0;    s.push(0);    while(!s.empty())    {        int u=s.front();        vis[u]=0;        s.pop();        for(int i=0;i<n;i++)        {            if(a[u][i]!='0')            {                int v=i;                int w=a[u][i]-'0';                if(dist[v]>dist[u]+w)                {                    dist[v]=dist[u]+w;                    if(vis[v]==0)                    {                        vis[v]=1;                        s.push(v);                    }                }            }        }    }}int main(){    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)scanf("%s",a[i]);        SPFA();        memset(ans,0,sizeof(ans));        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {                if(a[i][j]=='0')continue;                else                {                    int w=a[i][j]-'0';                    if(dist[i]+w==dist[j])                    {                        ans[j]++;                    }                }            }        }        __int64 output=1;        for(int i=1;i<n;i++)        {            output*=(__int64)ans[i];            output%=1000000000+7;        }        printf("%I64d\n",output);    }}