zoukankan      html  css  js  c++  java
  • light oj 1019【最短路模板】

    1019 - Brush (V)
    Time Limit: 2 second(s) Memory Limit: 32 MB

    Tanvir returned home from the contest and got angry after seeing his room dusty. Who likes to see a dusty room after a brain storming programming contest? After checking a bit he found that there is no brush in him room. So, he called Atiq to get a brush. But as usual Atiq refused to come. So, Tanvir decided to go to Atiq's house.

    The city they live in is divided by some junctions. The junctions are connected by two way roads. They live in different junctions. And they can go to one junction to other by using the roads only.

    Now you are given the map of the city and the distances of the roads. You have to find the minimum distance Tanvir has to travel to reach Atiq's house.

    Input

    Input starts with an integer T (≤ 100), denoting the number of test cases.

    Each case starts with a blank line. The next line contains two integers N (2 ≤ N ≤ 100) and M (0 ≤ M ≤ 1000), means that there are N junctions and M two way roads. Each of the next M lines will contain three integers u v w (1 ≤ u, v ≤ N, w ≤ 1000), it means that there is a road between junction u and v and the distance is w. You can assume that Tanvir lives in the 1st junction and Atiq lives in the Nth junction. There can be multiple roads between same pair of junctions.

    Output

    For each case print the case number and the minimum distance Tanvir has to travel to reach Atiq's house. If it's impossible, then print 'Impossible'.

    Sample Input

    Output for Sample Input

    2

    3 2

    1 2 50

    2 3 10

    3 1

    1 2 40

    Case 1: 60

    Case 2: Impossible

    练练模板  

    #include<stdio.h>
    #include<string.h>
    #define MAX 1010
    #define INF 0x3f3f3f
    int n,m;
    int vis[MAX],dis[MAX];
    int map[MAX][MAX];
    void init()
    {
    	int i,j;
    	for(i=1;i<=n;i++)
    	    for(j=1;j<=n;j++)
    	        map[i][j]=i==j?0:INF;
    }
    void dijktra()
    {
    	int i,j,next,min;
    	for(i=1;i<=n;i++)
    	    dis[i]=map[1][i];
    	memset(vis,0,sizeof(vis));
    	vis[1]=1;
    	next=1;
    	for(i=2;i<=n;i++)
    	{
    		min=INF;
    		for(j=1;j<=n;j++)
    		{
    			if(!vis[j]&&min>dis[j])
    			{
    				next=j;
    				min=dis[j];
    			}
    		}
    		vis[next]=1;
    		for(j=1;j<=n;j++)
    		{
    			if(!vis[j]&&dis[j]>dis[next]+map[next][j])
    			    dis[j]=dis[next]+map[next][j];
    		} 
    	}
    	if(dis[n]==INF)
    	printf("impossible
    ");
    	else
    	printf("%d
    ",dis[n]);
    }
    int main()
    {
    	int k,t,j,i,a,b,c;
    	scanf("%d",&t);
    	k=0;
    	while(t--)
    	{
    		scanf("%d%d",&n,&m);
    		init();
    		while(m--)
    		{
    			scanf("%d%d%d",&a,&b,&c);
    			if(map[a][b]>c)
    			    map[a][b]=map[b][a]=c;
    		}
    		printf("Case %d:",++k);
    		dijktra();
    	}
    	return 0;
    } 
    

      

  • 相关阅读:
    vue-cli创建的webpack工程中引用ExtractTextPlugin导致css背景图设置无效的解决方法
    如何做一个技术全面的架构师
    如何做一个技术全面的架构师
    最佳实践:阿里云VPC、ECS支持IPv6啦!
    最佳实践:阿里云VPC、ECS支持IPv6啦!
    Windows10中打开git bash闪退解决方案
    webpack打包多入口配置
    Express 文档(常见问题)
    Express 文档(常见问题)
    bzoj1089严格n元树
  • 原文地址:https://www.cnblogs.com/tonghao/p/4811313.html
Copyright © 2011-2022 走看看