POJ 3723 Conscription (最小生成树)

题意：给定 n 个男人，m 个女人，和 r 个男女之间的关系，每个征募一个人要用10000元，但是如果有关系可以少花一些钱，即10000-亲密度，

求一个最小要花多少钱。

析：最后生成的关系肯定是一片森林，也就是最大权森林，但是我可以把权值取反，然后就是一个求最小森林了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10000 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

struct Node{
  int u, v, val;
  Node() { }
  Node(int uu, int vv, int va) : u(uu), v(vv), val(va) { }
  bool operator < (const Node &p) const{
    return val < p.val;
  }
};
Node a[maxn*5];
int p[maxn*2];
int Find(int x){ return x == p[x] ? x : p[x] = Find(p[x]); }

int solve(int r){
  sort(a, a + r);
  int ans = 0;
  for(int i = 0; i < r; ++i){
    int x = Find(a[i].u);
    int y = Find(a[i].v);
    if(x != y) p[y] = x, ans += a[i].val;
  }
  return ans;
}

int main(){
  int T;  cin >> T;
  while(T--){
    int r;
    scanf("%d %d %d", &n, &m, &r);
    for(int i = 0; i < n+m; ++i)  p[i] = i;
    for(int i = 0; i < r; ++i){
      int u, v, val;
      scanf("%d %d %d", &u, &v, &val);
      a[i] = Node(u, v+n, -val);
    }
    printf("%d\n", 10000 * (n+m) + solve(r));
  }
  return 0;
}