zoukankan      html  css  js  c++  java
  • Choose the best route

    Choose the best route

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 10482    Accepted Submission(s): 3373


    Problem Description
    One day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her friend’s home as soon as possible . Now give you a map of the city’s traffic route, and the stations which are near Kiki’s home so that she can take. You may suppose Kiki can change the bus at any station. Please find out the least time Kiki needs to spend. To make it easy, if the city have n bus stations ,the stations will been expressed as an integer 1,2,3…n.
     

    Input
    There are several test cases.
    Each case begins with three integers n, m and s,(n<1000,m<20000,1=<s<=n) n stands for the number of bus stations in this city and m stands for the number of directed ways between bus stations .(Maybe there are several ways between two bus stations .) s stands for the bus station that near Kiki’s friend’s home.
    Then follow m lines ,each line contains three integers p , q , t (0<t<=1000). means from station p to station q there is a way and it will costs t minutes .
    Then a line with an integer w(0<w<n), means the number of stations Kiki can take at the beginning. Then follows w integers stands for these stations.
     

    Output
    The output contains one line for each data set : the least time Kiki needs to spend ,if it’s impossible to find such a route ,just output “-1”.
     

    Sample Input
    5 8 5 1 2 2 1 5 3 1 3 4 2 4 7 2 5 6 2 3 5 3 5 1 4 5 1 2 2 3 4 3 4 1 2 3 1 3 4 2 3 2 1 1
     

    Sample Output
    1 -1
     

    Author
    dandelion
     

    Source
     

    Recommend
    lcy   |   We have carefully selected several similar problems for you:  2112 1217 1142 1385 2923 
     
     
     
    #include<stdio.h>
    #include<string.h>
    #define INF 0xfffffff
    int map[1010][1010],v[1010],n,dis[1010];
    void dijs(int x)
    {
        int i,j,k,min;
        for(i=0;i<=n;i++)
        {
            dis[i]=map[x][i];
            //printf("%d
    ",dis[i]);
        }
        v[x]=1;
        for(i=1;i<=n;i++)
        {
            k=-1;
            min=INF;
            for(j=1;j<=n;j++)
            {
                if(!v[j]&&min>dis[j])
                {
                    k=j;
                    min=dis[j];
                }
            }
            if(k==-1)
                break;
            v[k]=1;
            for(j=1;j<=n;j++)
            {
                if(!v[j]&&dis[j]>dis[k]+map[k][j])
                {
                    dis[j]=dis[k]+map[k][j];
                }
            }
        }
    }
    int main()
    {
        int m,ans,num;
        while(scanf("%d%d%d",&n,&m,&ans)!=EOF)
        {
            int i,j,min;
            for(i=0;i<=n;i++)
            for(j=0;j<=n;j++)
                map[i][j]=INF;
            while(m--)
            {
                int a,b,c;
                scanf("%d%d%d",&a,&b,&c);
                if(map[b][a]>c)
                    map[b][a]=c;
            }
            memset(v,0,sizeof(v));
            dijs(ans);
            min=INF;
            scanf("%d",&num);
            while(num--)
            {
                int st;
                scanf("%d",&st);
                if(dis[st]<min)
                min=dis[st];
            }
            if(min<INF)
                printf("%d
    ",min);
            else
                printf("-1
    ");
        }
    }


    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #include<queue>
    using namespace std;
    #define INF 0x3f3f3f
    #define min(a,b)(a>b?b:a)
    int head[21010],dis[21010],vis[21010],m,n,s,cnt;
    struct node 
    {
    	int u,v;
    	int val,next;
    }edge[21010];
    void add(int u,int v,int val)
    {
    	edge[cnt].u=u;
    	edge[cnt].v=v;
    	edge[cnt].val=val;
    	edge[cnt].next=head[u];
    	head[u]=cnt++;
    }
    void SPFA()
    {
    	queue<int>q;
    	q.push(s);
    	memset(vis,0,sizeof(vis));
    	memset(dis,INF,sizeof(dis));
    	vis[s]=1;
    	dis[s]=0;
    	while(!q.empty())
    	{
    		int u=q.front();
    		q.pop();
    		vis[u]=0;
    		for(int i=head[u];i!=-1;i=edge[i].next)
    		{
    			int v=edge[i].v;
    			if(dis[v]>dis[u]+edge[i].val)
    			{
    				dis[v]=dis[u]+edge[i].val;
    				if(!vis[v])
    				{
    					vis[v]=1;
    					q.push(v);
    				}
    			}
    		}
    	}
    }
    int main()
    {
    	while(scanf("%d%d%d",&m,&n,&s)!=EOF)
    	{
    		memset(head,-1,sizeof(head));
    		int a,b,c;
    		cnt=0;
    		while(n--)
    		{
    			scanf("%d%d%d",&a,&b,&c);
    			add(a,b,c);
    			add(b,a,c);
    		}
    		SPFA();
    		scanf("%d",&c);
    		int minn=INF;
    		while(c--)
    		{
    			scanf("%d",&a);
    			minn=min(minn,dis[a]);
    		}
    		if(minn<INF)
    		printf("%d
    ",minn);
    		else
    		printf("-1
    ");
    	}
    	return 0;
    }


  • 相关阅读:
    SQL中的字符串字段根据某字段实现自增
    SQL中的字符串字段实现自增
    ECS Windows服务器IIS FTP登陆提示“530 valid hostname is expected”
    HTML中动态生成内容的事件绑定问题
    JavaScript对JSON数据进行排序和搜索
    IIS站点下多应用程序 C#获取根目录方法
    asp.net中form表单多个按钮submit提交到后台的实例
    C#从一个SqlCommand对象生成可执行的SQL语句
    传递参数
    js创建jsonArray传输至后台及后台解析
  • 原文地址:https://www.cnblogs.com/playboy307/p/5273821.html
Copyright © 2011-2022 走看看