poj 1062 (dij最短路径)

Description

Input

Output

Sample Input

1 4
10000 3 2
2 8000
3 5000
1000 2 1
4 200
3000 2 1
4 200
50 2 0

Sample Output

5250

Source

浙江

AC代码：

#include<iostream>
using namespace std;
int m,n;
int dis[105];
int G[105][105];
int p[105],r[105];
void dij(int st,int ed){
    int vis[105];
    for(int i=2;i<=n;i++){
        vis[i]=0;
        dis[i]=G[1][i];
        if(r[i]<st || r[i]>ed){
            vis[i]=1;
            dis[i]=999999;
        }
    }
    dis[1]=0;
    vis[1]=1;
    for(int i=1;i<n;i++){
        int v,min=999999;
        for(int j=1;j<=n;j++){
            if(!vis[j] && dis[j]<min){
                min=dis[j];
                v=j;
            }
        }
        if(min==999999)
            break;
        vis[v]=1;
        for(int j=1;j<=n;j++){
            if(!vis[j] && dis[j]>min+G[v][j])
                dis[j]=min+G[v][j];
        }
    }
}
int main(){
    while(cin>>m>>n){
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                if(i==j)
                    G[i][j]=0;
                else
                    G[i][j]=999999;
        for(int i=1;i<=n;i++){
            int t;
            cin>>p[i]>>r[i]>>t;
            for(int j=1;j<=t;j++){
                int v,price;
                cin>>v>>price;
                G[i][v]=price;
            }
        }
        int ans=999999;
        for(int i=r[1]-m;i<=r[1];i++){     
            dij(i,i+m);                    //枚举等级范围
            for(int j=1;j<=n;j++){
                if(ans>dis[j]+p[j]){
                    ans=dis[j]+p[j];
                }
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

poj 1062 (dij最短路径),布布扣,bubuko.com

poj 1062 (dij最短路径)

原文：http://blog.csdn.net/my_acm/article/details/38011213