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最新推荐文章于 2019-07-14 13:24:44 发布

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#最短路 #杭电

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本文介绍了一个寻找从城市到海滨最短路径的问题，通过Floyd算法解决了该问题。输入包含多个测试用例，每个用例包括城镇数量及城镇间的连接关系，最终输出从起点到达海滨城镇的最短距离。

Seaside

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1562 Accepted Submission(s): 1138

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

Sample Output

Source

2010 Asia Regional Harbin

Recommend

lcy

数据较小，floyd算法简洁方便。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int N = 15;
int map[N][N];
int vis[N];
int n;
void floyd() {
	for(int k=0; k<n; k++) 
	{
		for(int i=0; i<n; i++) 
		{
			for(int j=0; j<n; j++)
			{
				map[i][j]=min(map[i][j],map[i][k]+map[k][j]);
			}
		}
	}
}
int main() {
	while(scanf("%d",&n)!=EOF) {
		memset(vis,0,sizeof(vis)); 
		for(int i=0; i<n; i++) {
			for(int j=0; j<n; j++) {
				map[i][j]=i==j?0:inf;
			}
		}
		for(int l=0; l<n; l++) {
			int m,a,b;
			scanf("%d%d",&m,&vis[l]);
			for(int j=0; j<m; j++) {
				scanf("%d%d",&a,&b);
				map[l][a]=b;
			}
		}
		floyd();
		int minn=inf;
		for(int l=1; l<n; l++) {
			if(vis[l])
			minn=min(minn,map[0][l]);
		}
		printf("%d\n",minn);
	}
	return 0;
}