zoukankan      html  css  js  c++  java
  • poj3723_Conscription

    Conscription
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 12393   Accepted: 4350

    Description

    Windy has a country, and he wants to build an army to protect his country. He has picked up N girls and M boys and wants to collect them to be his soldiers. To collect a soldier without any privilege, he must pay 10000 RMB. There are some relationships between girls and boys and Windy can use these relationships to reduce his cost. If girl x and boy y have a relationship d and one of them has been collected, Windy can collect the other one with 10000-d RMB. Now given all the relationships between girls and boys, your assignment is to find the least amount of money Windy has to pay. Notice that only one relationship can be used when collecting one soldier.

    Input

    The first line of input is the number of test case.
    The first line of each test case contains three integers, NM and R.
    Then R lines followed, each contains three integers xiyi and di.
    There is a blank line before each test case.

    1 ≤ NM ≤ 10000
    0 ≤ R ≤ 50,000
    0 ≤ xi < N
    0 ≤ yi < M
    0 < di < 10000

    Output

    For each test case output the answer in a single line.

    Sample Input

    2
    
    5 5 8
    4 3 6831
    1 3 4583
    0 0 6592
    0 1 3063
    3 3 4975
    1 3 2049
    4 2 2104
    2 2 781
    
    5 5 10
    2 4 9820
    3 2 6236
    3 1 8864
    2 4 8326
    2 0 5156
    2 0 1463
    4 1 2439
    0 4 4373
    3 4 8889
    2 4 3133
    

    Sample Output

    71071
    54223
    

    Source

    [Submit]   [Go Back]   [Status]   [Discuss]

    刚开始差点理解错意思,原来是男女两种人,windy必须要买掉所有人是(N+M)*10000,然后每两个人认识会给windy减少d的花费,kruskal中按照从大到小排序即可。

    另外也可以把-d存进去,那样kruskal中就不用修改了。

    #include <iostream>
    #include <cstdio>
    #include <vector>
    #include <queue>
    #include <cstring>
    #include <algorithm>
    #include <cstdlib>
    #define for(i,x,n) for(int i=x;i<n;i++)
    #define ll long long int
    #define INF 0x3f3f3f3f
    #define MOD 1000000007
    #define MAX_N 50005
    
    using namespace std;
    
    struct edge{int u,v,cost;};
    edge es[MAX_N];
    int V,E;
    
    int par[MAX_N];
    int depth[MAX_N];//记录每个节点下面到位深度
    
    void init(int n)//初始化
        {for(i,0,n) {par[i]=i;depth[i]=1;}}
    int findf(int t)//寻找根节点
        { return t==par[t] ? t:par[t]=findf(par[t]);}
    bool same(int x,int y)//是否在同一组内
        {return findf(x)==findf(y);}
    void unite(int t1,int t2){//成为一组
        int f1=findf(t1);
        int f2=findf(t2);
        if(f1==f2){
            return ;
        }
        if(depth[f1]<depth[f2]){
            par[f1]=f2;
        }else{
            par[f2]=f1;
            if(depth[f1]==depth[f2]){
                depth[f1]++;//记录深度
            }
        }
    }
    
    bool comp(edge a,edge b){
        return a.cost>b.cost;
    }
    
    int kruskal(){
        sort(es,es+E,comp);
        init(V);
        int res=0;
        for(i,0,E){
            edge e=es[i];
            if(!same(e.u,e.v)){
                unite(e.u,e.v);
                res+=e.cost;
            }
        }
        return res;
    }
    
    int main()
    {
        //freopen("data.txt", "r", stdin);
        //freopen("data.out", "w", stdout);
        int N,M,R;
        int x,y,d;
        int n;
        scanf("%d",&n);
        while(n--){
            scanf("%d %d %d",&N,&M,&R);
            V=N+M;
            E=R;
            for(i,0,R){
                scanf("%d %d %d",&x,&y,&d);
                es[i].u=x;
                es[i].v=y+N;
                es[i].cost=d;
            }
            int res=kruskal();
            printf("%d
    ",(N+M)*10000-res);
    
        }
        //fclose(stdin);
        //fclose(stdout);
        return 0;
    }
  • 相关阅读:
    引用jar版本错误
    一次皆大欢喜的程序员跳槽加薪
    Netty
    JAVA笔记-如何将百万级数据高效的导出到Excel表单
    100个高质量Java开发者博客
    Stackoverflow上人气最旺的10个Java问题
    偏执却管用的10条Java编程技巧
    如何生成每秒百万级别的 HTTP 请求?
    Google浏览器Chrome安装失败,错误代码0xa0430721解决办法
    Redis介绍及常用命令
  • 原文地址:https://www.cnblogs.com/TWS-YIFEI/p/6359037.html
Copyright © 2011-2022 走看看