hdu 6026 Deleting Edges 江苏徐州邀请赛K

Deleting Edges

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1693    Accepted Submission(s): 575

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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#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
string s[100];
int mp[100][100],n;
int dis[100],vis[100],in_[100];
struct node{
    int to,v;
    friend bool operator < (node a,node b){return a.v<b.v;};
    friend bool operator > (node a,node b){return a.v>b.v;};
};
void dij(){
    priority_queue<node,vector<node>,greater<node> >q;
    memset(dis,0x3f,sizeof(dis));
    memset(vis,0,sizeof(vis));
    dis[0]=0;vis[0]=1;q.push({0,0});
    while(!q.empty()){
        node u=q.top();q.pop();
        if(u.v>dis[u.to]) continue;
        for(int i=0;i<n;i++){
            if(mp[u.to][i]==0) continue;
            if(dis[i]>dis[u.to]+mp[u.to][i]){
                dis[i]=dis[u.to]+mp[u.to][i];
                q.push({i,dis[i]});
            }
        }
    }
}
int main(){
    while(cin>>n){
        for(int i=0;i<n;i++) cin>>s[i];
        memset(in_,0,sizeof(in_));
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                mp[i][j]=s[i][j]-'0';
        dij();
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(mp[i][j]==0) continue;
                if(dis[i]+mp[i][j]==dis[j]) in_[j]++;
            }
        }
        long long ans=1;
        for(int i=1;i<n;i++){
            if(in_[i])(ans*=in_[i])%=mod;
        }
        printf("%lld\n",ans);
    }
    return 0;
} 

View Code

#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
string s[100];
int mp[100][100],n;
int dis[100],vis[100],in_[100];
void dij(){
    //priority_queue<int,vector<int>,greater<int> >q;
    queue<int>q;
    memset(dis,0x3f,sizeof(dis));
    memset(vis,0,sizeof(vis));
    dis[0]=0;vis[0]=1;q.push(0);
    while(!q.empty()){
        int u=q.front();q.pop();vis[u]=0;
        for(int i=0;i<n;i++){
            if(mp[u][i]==0) continue;
            if(dis[i]>dis[u]+mp[u][i]){
                dis[i]=dis[u]+mp[u][i];
                if(!vis[i]){
                    q.push(i);
                    vis[i]=1;
                }
            }
        }
    }
}
int main(){
    while(cin>>n){
        for(int i=0;i<n;i++) cin>>s[i];
        memset(in_,0,sizeof(in_));
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                mp[i][j]=s[i][j]-'0';
        dij();
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(mp[i][j]==0) continue;
                if(dis[i]+mp[i][j]==dis[j]) in_[j]++;
            }
        }
        long long ans=1;
        for(int i=1;i<n;i++){
            if(in_[i])(ans*=in_[i])%=mod;
        }
        printf("%lld\n",ans);
    }
    return 0;
} 

View Code