HDU 3665 Seaside （最短路--floyd）

Seaside

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1528    Accepted Submission(s): 1107

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

 

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题意;现在在小镇0，求到海边的最短距离。

输入n代表n个城镇，输入Mi   Pi ，pi=1表示城镇i在海边Pi=0表示没有在海边，在输入Mi行SMi  Lmi 表示城镇i与城镇Smi相连距离为 Lmi  

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define INF 0xfffffff
int map[15][15];
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int i,j,k,l,from,to,p,num;
		for(i=0;i<=n;i++)
		for(j=0;j<=n;j++)
		{
			map[i][j]=INF;
			if(i==j)
			map[i][j]=0;
		}
		for(from=0;from<n;from++)
		{
			scanf("%d%d",&p,&num);
			if(num)
			map[from][n]=0;
			for(j=0;j<p;j++)
			{
				scanf("%d%d",&to,&l);
				map[from][to]=l;
			}
		}
		for(k=0;k<=n;k++)
		for(i=0;i<=n;i++)
		for(j=0;j<=n;j++)
		{
			if(map[i][j]>map[i][k]+map[k][j])
			map[i][j]=map[i][k]+map[k][j];
		}
		printf("%d\n",map[0][n]);
	}
	return 0;
}