hdu 3665Seaside(简单floyd)

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


Problem DescriptionXiaoY 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.
InputThere 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.
OutputEach case takes one line, print the shortest length that XiaoY reach seaside.
Sample Input

51 01 12 02 33 11 14 1000 10 1

Sample Output

2

Source2010 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 1e12using 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;}