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 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 10

9+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

題目大意：

給你N個點，然後給你一個鄰接矩陣，讓你求一共有多少種刪除邊的方式，使得構成一個生成樹之後，從0到任意點v的距離都是原圖的最短路的距離。

思路：

首先我們預處理出從點0到其他各個點的最短路的距離dist【i】.

那麼對應如果有dist【i】+w==dist【j】，那麼從i到j這條邊就可以作為構成生成樹的一條邊。

那麼我們將所有可以未做生成樹的邊的個數處理出來，那麼對應在每個點上邊取一條邊出來作為答案即可。

那麼累成結果就行了。

Ac程式碼：

#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);
    }
}