Conscription POJ - 3723

最新推荐文章于 2022-05-19 23:32:53 发布

原创最新推荐文章于 2022-05-19 23:32:53 发布 · 187 阅读

0 ·

CC 4.0 BY-SA版权

专题六最小生成树专栏收录该内容

5 篇文章

订阅专栏

博客围绕征兵问题展开，Windy要组建军队，招募N个女孩和M个男孩，无特权招募需每人10000元，可利用男女关系降低成本。给定男女关系及相关数据，要求找出最小花费，还给出了输入输出格式及示例，最后有AC代码。

Conscription

POJ - 3723

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, N, M and R.
Then R lines followed, each contains three integers xi, yi and di.
There is a blank line before each test case.

1 ≤ N, M ≤ 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

Sample Output

71071
54223

AC代码

//#include<bits/stdc++.h>
#define _CRT_SBCURE_NO_DEPRECATE
#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
//#define UP(i,x,y) for(int i=x;i<=y;i++)
//#define DOWN(i,x,y) for(int i=x;i>=y;i--)
#define sdddd(x,y,z,k) scanf("%d%d%d%d", &x, &y, &z, &k)
#define sddd(x,y,z) scanf("%d%d%d", &x, &y, &z)
#define sdd(x,y) scanf("%d%d", &x, &y)
#define sd(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define lson k<<1
#define rson k<<1|1
#define mid (l+r)/2
#define ms(x, y) memset(x, y, sizeof x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MOD 1000000007
const int maxn = 10050;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
int n, m, s, t;
int len;
struct road
{
    int u, v, cost;
};
road G[maxn*5];
bool vis[maxn];
int par[maxn<<1];
void init()
{
    for(int i = 0; i < maxn<<1; i++) par[i] = i;
}

int Find(int x)
{
    return par[x]==x? x : par[x] = Find(par[x]);
}
bool same(int x, int y)
{
    int fx = Find(x), fy = Find(y);
    if(fx == fy)return true;
    else        return false;
}
void unite(int x, int y)
{
    int fx = Find(x), fy = Find(y);
    if(fx == fy) return;
    else{
        par[fx] = fy;
        return;
    }
}
bool cmp(road a, road b)
{
    return a.cost>b.cost;
}
int kruskal()
{
    sort(G, G+len, cmp);
    init();
    int res = 0;
    for(int i = 0; i < len; i++)
    {
        road e = G[i];
        if(!same(e.u, e.v))
        {
            unite(e.u, e.v);
            res+=e.cost;
        }
    }
    return res;
}
int main()
{
	//freopen("out.txt", "w", stdout);
	sd(t);
	while(t--)
    {
        ms(G,0);
        sddd(n,m,s);
        len = 0;
        int ta, tb, tc;
        for(int i = 0; i < s; i++)
        {
            sddd(ta, tb, tc);
            G[len].u = ta;
            G[len].v = tb+n;
            G[len++].cost = tc;
        }
        int ans = (n+m)*10000-kruskal();
        printf("%d\n", ans);
    }
	return 0;
}