2017中国大学生程序设计竞赛 - 女生专场 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): 567    Accepted Submission(s): 210

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  nodes, labeled from 0 to . 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  edges.
(2) For every vertice , the distance between 0 and  on the tree is equal to the length of shortest path from 0 to  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  and , 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 .

 

Input

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

 

Output

For each test case, print a single line containing a single integer, denoting the answer modulo .

 

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

题意：给你一个图，让你删掉一些边后变成一棵树，这棵树的根是0号节点，要满足在树上从0到任意节点的最短距离和原图相等。问总共有多少种删法？

做法：

用dijkstra求出原图中0点到每个点的最短路的长度，再暴力的跑一遍判断能否通过别的点同样使得0点到当前点的距离仍是最短路径，如果有一个可替代点，则表示对于当前点存在一种可以删边的方法，将所有的可能删的方法相乘即可。

#include <iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<deque>
#include<vector>
#define ll long long
#define inf 0x3f3f3f3f
#define mod 1000000007;
using namespace std;
int n;
int e[][];
int d[];
bool book[];
void dij()
{
    memset(d,inf,sizeof(d));
    memset(book,,sizeof(book));
    for(int i=;i<=n;i++) d[i]=e[][i];
    book[]=;
    d[]=;
    int k=-;
    int mi=inf;
    while()
    {
        k=-;
        mi=inf;
        for(int i=;i<=n;i++)
        {
            if(!book[i]&&d[i]<mi)
            {
                mi=d[i];
                k=i;
            }
        }
        if(k==-) break;
        book[k]=;
        for(int i=;i<=n;i++)
        {
            if(!book[i]&&d[i]>d[k]+e[k][i])
            {
                d[i]=d[k]+e[k][i];
            }
        }
    }
}
int main()
{
    while(~scanf("%d",&n))
    {
        char s[];
        for(int i=;i<=n;i++)
        {
            cin>>s;
            for(int j=;j<n;j++)
            {
                if(s[j]=='')
                {
                    e[i][j+]=inf;
                }
                else e[i][j+]=s[j]-'';
            }
        }
        dij();
        ll ans=;
        ll temp=;
        for(int i=;i<=n;i++)
        {
            temp=;
            for(int j=;j<=n;j++)
            {
                if(d[i]==d[j]+e[i][j])
                {
                    temp++;
                }
            }
            ans=(ans*temp)%mod;
        }
        printf("%lld\n",ans);
    }
    return ;
}