zoukankan      html  css  js  c++  java
  • poj 3159 dijkstra 最短路

    Description

    During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought the kids of flymouse’s class a large bag of candies and had flymouse distribute them. All the kids loved candies very much and often compared the numbers of candies they got with others. A kid A could had the idea that though it might be the case that another kid B was better than him in some aspect and therefore had a reason for deserving more candies than he did, he should never get a certain number of candies fewer than B did no matter how many candies he actually got, otherwise he would feel dissatisfied and go to the head-teacher to complain about flymouse’s biased distribution.

    snoopy shared class with flymouse at that time. flymouse always compared the number of his candies with that of snoopy’s. He wanted to make the difference between the numbers as large as possible while keeping every kid satisfied. Now he had just got another bag of candies from the head-teacher, what was the largest difference he could make out of it?

    Input

    The input contains a single test cases. The test cases starts with a line with two integers N and M not exceeding 30 000 and 150 000 respectively. N is the number of kids in the class and the kids were numbered 1 through N. snoopy and flymouse were always numbered 1 and N. Then follow M lines each holding three integers A, B and c in order, meaning that kid A believed that kid B should never get over c candies more than he did.

    Output

    Output one line with only the largest difference desired. The difference is guaranteed to be finite.

    Sample Input

    2 2
    1 2 5
    2 1 4

    Sample Output

    5


    分析题意:属于单源最短路
    dijkstra+优先队列 代码如下
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<algorithm>
    #include<queue>
    #include<stack>
    #include<vector>
    #include<map>
    using namespace std;
    struct node
    {
        int e;
        int w;
        bool friend operator <(node s1,node s2)
        {
            return s1.w>s2.w;
        }
    };
    vector<vector<node> > edge;
    priority_queue<node>q;
    bool beused[30005];
    node ww;
    int main()
    {
        int n,m;
        int A,B,C;
        scanf("%d%d",&n,&m);
        edge.clear();
        edge.resize(n+1);
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d",&A,&B,&C);
            ww.e=B;
            ww.w=C;
            edge[A].push_back(ww);
        }
        memset(beused,0,sizeof(beused));
        while(!q.empty())
            q.pop();
        ww.e=1;
        ww.w=0;
        q.push(ww);
        while(!q.empty())
        {
            ww=q.top();
            q.pop();
            if(beused[ww.e])
                continue;
            beused[ww.e]=true;
            if(ww.e==n)
                break;
            int j=edge[ww.e].size();
            for(int i=0;i<j;i++)
            {
                node r;
               r.e=edge[ww.e][i].e;
               if(beused[r.e])
                continue;
                r.w=ww.w+edge[ww.e][i].w;
                if(!beused[r.e])
                q.push(r);
            }
        }
         printf("%d
    ",ww.w);
       return 0;
    }
    
     
  • 相关阅读:
    idea 安装lombok 插件过程
    typora快捷键之速成笔记
    Android端抓取日志
    接口测试Case之面向页面对象编写及规范
    Jenkins + jmeter + ant + git 自动化集成
    自动化测试使用流程
    Jenkins git 的配置及问题解决
    传智springMVC笔记
    session 原理 数据结构
    Redis 3.0 集群搭建
  • 原文地址:https://www.cnblogs.com/hsd-/p/4681641.html
Copyright © 2011-2022 走看看