Deleting Edges

                                                                                 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
                                                                                                                 Total Submission(s): 18    Accepted Submission(s): 9


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 nodes i 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 the j-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 modulo 109+7.
 

Sample Input
2
01
10
4
0123
1012
2101
3210
 

Sample Output
1
6
 

Source
2017中国大学生程序设计竞赛 - 女生专场


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题目的意思是给出一张图，求最短路数的个数有多少种
思路：求出最短路，在求最短路时记录到达每个点的方案数，结果就是所有点方案数之和

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;
const int mod=1e9+7;

int n,m;
int mp[55][55];
int dis[55];
int vis[55];
LL cnt[55];
struct node
{
    int id,val;
    friend bool operator<(node a,node b)
    {
        return a.val>b.val;
    }
};

void djstl(int o)
{
    memset(dis,INF,sizeof dis);
    memset(vis,0,sizeof vis);
    memset(cnt,0,sizeof cnt);
    node f,d;
    f.id=o,f.val=0;
    priority_queue<node>q;
    q.push(f);
    vis[o]=cnt[o]=1,dis[o]=0;
    while(!q.empty())
    {
        f=q.top();
        q.pop();
        vis[f.id]=1;
        for(int i=0; i<n; i++)
        {
            if(!vis[i]&&f.val+mp[f.id][i]<dis[i])
            {
                dis[i]=f.val+mp[f.id][i];
                cnt[i]=1;
                d.id=i;
                d.val=dis[i];
                q.push(d);
            }
            else if(!vis[i]&&f.val+mp[f.id][i]==dis[i])
            {
                cnt[i]++;
            }
        }
    }
}


int main()
{
    char s[55];
    while(~scanf("%d",&n))
    {
        memset(mp,INF,sizeof mp);
        for(int i=0; i<n; i++)
        {
            scanf("%s",&s);
            for(int j=0; j<n; j++)
                if(s[j]!='0')
                    mp[i][j]=s[j]-'0';
        }

        djstl(0);
        LL ans=1;
        for(int i=0; i<n; i++)
        {
            ans*=cnt[i];
            ans%=mod;
        }
        printf("%d\n",ans);
    }
    return 0;
}