zoukankan      html  css  js  c++  java
  • hdoj_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): 4888    Accepted Submission(s): 1547


    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
    注意题目是有向图,将起点终点逆置,就能用dijkstra了

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<map>
    #include<cmath>
    #include<vector>
    #include<algorithm>
    #include<set>
    #include<string>
    #include<queue>
    #include <stack>
    using namespace std;
    #pragma warning(disable : 4996)
    
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<map>
    #include<cmath>
    #include<vector>
    #include<algorithm>
    #include<set>
    #include<string>
    #include<queue>
    #include <stack>
    using namespace std;
    #pragma warning(disable : 4996)
    
    const int MAXN = 1005;
    const int INF = 999999;
    int n;
    int maps[MAXN][MAXN];
    bool visited[MAXN];
    int dist[MAXN];
    
    void Dijkstra(int s)
    {
    	int i, j;
    	int minValue, minNode;
    
    	dist[s] = 0;
    	visited[s] = true;
    	for (i = 1; i <= n; i++)
    	{
    		dist[i] = maps[s][i];
    	}
    	for (i = 1; i <= n; i++)
    	{
    		minValue = INF;
    		minNode = 0;
    		for (j = 1; j <= n; j++)
    		{
    			if(!visited[j] && minValue > dist[j])
    			{
    				minNode = j;
    				minValue = dist[j];
    			}
    		}
    		if(minNode == 0)
    		{
    			break;
    		}
    		visited[minNode] = true;
    		for (j = 1; j <= n; j++)
    		{
    			if(!visited[j] && maps[minNode][j] != INF && dist[j] > dist[minNode] + maps[minNode][j])
    			{
    				dist[j] = dist[minNode] + maps[minNode][j];
    			}
    		}
    	}
    }
    
    int main()
    {
    	freopen("in.txt", "r", stdin);
    	int m, s, p, q, t, x, w, ans;
    	while (scanf("%d %d %d", &n, &m, &s) != EOF)
    	{
    		for (int i = 1; i <= n; i++)
    		{
    			for (int j = 1; j <= n; j++)
    			{
    				if(i == j)
    				{
    					maps[i][j] = 0;
    				}
    				else
    				{
    					maps[i][j] = INF;
    				}
    			}
    		}
    		while (m--)
    		{
    			scanf("%d %d %d", &p, &q, &t);
    			if(maps[q][p] > t)
    			{
    				maps[q][p] = t;
    			}
    		}
    		memset(visited, false, sizeof(visited));
    		Dijkstra(s);
    		ans = INF;
    		scanf("%d", &w);
    		while (w--)
    		{
    
    			scanf("%d", &x);
    			if(ans > dist[x])
    			{
    				ans = dist[x];
    			}
    		}
    		if(ans == INF)
    		{
    			printf("-1\n");
    		}
    		else
    		{
    			printf("%d\n", ans);
    		}
    	}
    	return 0;
    }



  • 相关阅读:
    zhcon安装过程记录
    diff和patch配合使用(转载备用)
    Linux /etc/passwd 和 /etc/group 文件格式
    APT 常用功能
    Linux邮件服务器架构
    出现segment fault 错误的几种原因
    位运算
    Linux启动过程中几个重要配置文件的执行过程
    2016-2017 ACM-ICPC, South Pacific Regional Contest (SPPC 16)
    Codeforces Round #439 (Div. 2) 题解
  • 原文地址:https://www.cnblogs.com/lgh1992314/p/5835041.html
Copyright © 2011-2022 走看看