HDU6026-Deleting Edges

Deleting Edges

                                                                                Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
                                                                                                                  Total Submission(s): 0    Accepted Submission(s): 0

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

2011040123101221013210

 

Sample Output

16

 

题意：有n个点，告诉你两两之间的边，让你去掉一些边，使得0号节点到其他节点都是最短路，问有多少种这样的图

解题思路：最短路，先建一个只包含最短路的有向无环图，每一个点选择任意一条入边即可生成一个树形图，那么树的种类就等于每个点的入度乘积

#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 longconst int INF=0x3f3f3f3f;#define mod 1000000007int n,s[60],nt[5009],e[5009],l[5009],visit[60],dis[60];LL a[60];char ch[60][60];struct node{    int id,l;    friend bool operator <(node a,node b)    {        return a.l>b.l;    }};void Dijkstra(){    memset(visit,0,sizeof visit);    memset(dis,INF,sizeof dis);    priority_queue<node>q;    node pre,nt1;    pre.id=0,pre.l=0,a[0]=1;    dis[0]=0;    q.push(pre);    while(!q.empty())    {        pre=q.top();        q.pop();        visit[pre.id]=1;        for(int i=s[pre.id];~i;i=nt[i])        {            int ee=e[i];            if(visit[ee]) continue;            if(dis[ee]>pre.l+l[i])            {                a[ee]=1;                dis[ee]=pre.l+l[i];                nt1.id=ee;                nt1.l=dis[ee];                q.push(nt1);            }            else if(dis[ee]==pre.l+l[i])                a[ee]++;        }    }    LL ans=1;    for(int i=0;i<n;i++)    {        ans*=a[i];        ans%=mod;    }    printf("%lld\n",ans);}int main(){    while(~scanf("%d",&n))    {        int cnt=0;        memset(s,-1,sizeof s);        for(int i=0;i<n;i++)        {            scanf("%s",ch[i]);            for(int j=0;j<n;j++)            {                if(ch[i][j]!='0')                    nt[cnt]=s[i],s[i]=cnt,e[cnt]=j,l[cnt++]=ch[i][j]-'0';            }        }        Dijkstra();    }    return 0;}