HDU6026 Deleting Edges

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;
}