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  • HDU 3665 Seaside

    Seaside

    http://acm.hdu.edu.cn/showproblem.php?pid=3665

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 802    Accepted Submission(s): 562

    Problem Description
    XiaoY is living in a big city, there are N towns in it and some towns near the sea. All these towns are numbered from 0 to N-1 and XiaoY lives in the town numbered ’0’. There are some directed roads connecting them. It is guaranteed that you can reach any town from the town numbered ’0’, but not all towns connect to each other by roads directly, and there is no ring in this city. One day, XiaoY want to go to the seaside, he asks you to help him find out the shortest way.
     
    Input
    There are several test cases. In each cases the first line contains an integer N (0<=N<=10), indicating the number of the towns. Then followed N blocks of data, in block-i there are two integers, Mi (0<=Mi<=N-1) and Pi, then Mi lines followed. Mi means there are Mi roads beginning with the i-th town. Pi indicates whether the i-th town is near to the sea, Pi=0 means No, Pi=1 means Yes. In next Mi lines, each line contains two integers SMi and LMi, which means that the distance between the i-th town and the SMi town is LMi.
     
    Output
    Each case takes one line, print the shortest length that XiaoY reach seaside.
     
    Sample Input
    5
    1 0
    1 1
    2 0
    2 3
    3 1
    1 1
    4 100
    0 1
    0 1
     
    Sample Output
     
    2
     
    Source
     
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    #include<stdio.h>
    
    #define INF 99999999
    
    int n,dis[15][15];
    int mark[15];
    
    void Floyed(){
        int i,j,k;
        for(k=0;k<n;k++)
            for(i=0;i<n;i++)
                for(j=0;j<n;j++)
                    if(dis[i][j]>dis[i][k]+dis[k][j])
                        dis[i][j]=dis[i][k]+dis[k][j];
    }
    
    int main(){
        while(scanf("%d",&n)!=EOF){
            int i,j;
            for(i=0;i<n;i++)
                for(j=0;j<n;j++)
                    if(i==j)
                        dis[i][j]=0;
                    else
                        dis[i][j]=INF;
            int m;
            int s,l;
            for(i=0;i<n;i++){
                scanf("%d%d",&m,&mark[i]);
                for(j=0;j<m;j++){
                    scanf("%d%d",&s,&l);
                    dis[i][s]=l;
                }
            }
            Floyed();
            int min=INF;
            for(i=0;i<n;i++)
                if(mark[i])
                    if(min>dis[0][i])
                        min=dis[0][i];
            printf("%d\n",min);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jackge/p/2839336.html
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