hdu 3665Seaside(简单floyd)

Seaside

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1006    Accepted Submission(s): 722

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 S
_Mi and L
_Mi, which means that the distance between the i-th town and the S
_Mi town is L
_Mi.

 

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

2010 Asia Regional Harbin

 

题目意思很好懂，当时我们竟然想到了一步步dp，范围很小，n最大才为10而已，可以直接用floyd，只是一直不敢交，觉得存在bug，因为n为0的时候不知道怎么处理。。最后突然交了一发，A了，应该没有n=0的数据。。

题目地址：Seaside

AC代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#define maxn 1e12
using namespace std;
long long dis[15][15];
int issea[15],n;

void floyd()
{
    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()
{
    int i,j;
    long long ans;
    while(cin>>n)
    {
        ans=maxn;
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
                dis[i][j]=maxn;
            dis[i][i]=0;
        }

        memset(issea,0,sizeof(issea));  //是否是海边
        for(i=0;i<n;i++)
        {
            int a,b,c,d;
            scanf("%d%d",&a,&b);
            issea[i]=b;
            for(j=0;j<a;j++)
            {
                scanf("%d%d",&c,&d);
                dis[i][c]=d;
            }
        }
        floyd();
        for(i=0;i<n;i++)
            if(issea[i]&&dis[0][i]<ans)
                ans=dis[0][i];
        cout<<ans<<endl;
    }
    return 0;
}