POJ 3723 Conscription

Conscription

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 1

Problem 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

<div><p>The first line of input is the number of test case.<br>The first line of each test case contains three integers, <i>N</i>, <i>M</i> and <i>R</i>.<br>Then <i>R</i> lines followed, each contains three integers <i>x_i</i>, <i>y_i</i> and <i>d_i</i>.<br>There is a blank line before each test case.<br></p><p>1 ≤ <i>N</i>, <i>M</i> ≤ 10000<br>0 ≤ <i>R</i> ≤ 50,000<br>0 ≤ <i>x_i</i> < <i>N</i><br>0 ≤ <i>y_i</i> < <i>M</i><br>0 < <i>d_i</i> < 10000<br></p>

 

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

 

题意

一共要招募n+m个士兵，原则上每招募一个士兵要花费10000元，但是假如在已经招募的士兵中存在与他关系亲密的人，那么就可以少花点钱，这种关系对一个士兵最多只能用一次，问最少可以花费多少钱把士兵招募齐。

题解

男女士兵存在亲密关系的话就可以少花点钱，找到一种招募的顺序，可以使亲密度最高，那么就可以节省最多的钱。把男士兵处理一下，避免和女士兵的编号重复了，这样就把问题转换成了最小生成树问题，记得把亲密度取反，求最小权值的时候。

#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<string>
#include<cctype>
#include<algorithm>
#include<vector>
using namespace std;
int par[30005];
int ranke[30005];
struct node
{
    int u, v, w;
};
vector<node> s;
bool cmp( node a, node b)
{
    return a.w < b.w;
}
int find(int x)
{
    if (par[x] == x) return x;
    else
        return par[x] = find(par[x]);
}
void unite(int x, int y)
{
    int a = find(x);
    int b = find(y);
    if (a == b) return;
    if (ranke[a]> ranke[b]) par[b] = a;
    else
    {
        par[a] = b;
        if (ranke[a] == ranke[b]) ranke[b]++;
    }
}
int main()
{
    int t;
    scanf("%d",&t);//cin又超时
    while (t--)
    {
        int n1, n2, m;
        scanf("%d %d %d", &n1, &n2, &m);
        int i;
        for (i = 0; i<30000; i++) { par[i] = i; ranke[i] = 0; }
        s.clear();
        for (i = 1; i <= m; i++)
        {
            node x;
            scanf("%d %d %d", &x.u, &x.v, &x.w);
            x.u += 10002;//为了与女区分
            x.w *= -1;//负处理
            s.push_back(x);
        }
        long long ans = (n1 + n2) * 10000;
        long long res = 0;
        sort(s.begin(), s.end(), cmp);
        for (i = 0; i <s.size(); i++)
        {
            node x = s[i];
            if (find(x.u) != find(x.v))
            {
                unite(x.u, x.v);
                res += x.w;
            }
        }
        printf("%lld\n\n", ans + res);
    }
    return 0;
}