POJ3723 Conscription 【并检查集合】

Conscription

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, N, M and R.

Then R lines followed, each contains three integers x_i, y_i and d_i.

There is a blank line before each test case.

1 ≤ N, M ≤ 10000

0 ≤ R ≤ 50,000

0 ≤ x_i < N

0 ≤ y_i < M

0 < d_i < 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

POJ Monthly Contest – 2009.04.05, windy7926778

题意：有N个女孩M个男孩去报名，每一个人报名花费10000，可是假设未报名的小孩跟已报名的小孩中有关系亲热的异性，那么能够少花一些钱。

给出若干男女关系之间的1~9999亲热度，报名费用为10000-（已报名的人中跟自己亲热度的最大值）。求全部人的报名费用和的最小值。

题解：不能由于看到男女关系就朝二分图想，实际上这题用的是最小生成树思想。虽然最后的结果可能是森林，仅仅需让ans+=森林个数*10000；

#include <stdio.h>
#include <string.h>
#include <algorithm>

#define maxn 20010
#define maxm 100010
#define COST 10000

int N, M, R, id;
int pre[maxn];
struct Node {
    int u, v, w;
} E[maxm];

bool cmp(Node a, Node b) {
    return a.w < b.w;
}

void addEdge(int u, int v, int w) {
    E[id].u = u; E[id].v = v; E[id++].w = w;
}

void getMap() {
    int x, y, d; id = 0;
    while(R--) {
        scanf("%d%d%d", &x, &y, &d);
        addEdge(x, y + N, COST - d);
    }
}

int ufind(int k) {
    int a = k, b;
    while(pre[k] != -1) k = pre[k];
    while(a != k) {
        b = pre[a];
        pre[a] = k;
        a = b;
    }
    return k;
}

bool same(int x, int y) {
    return ufind(x) == ufind(y);
}

void unite(int x, int y) {
    x = ufind(x);
    y = ufind(y);
    if(x != y) pre[y] = x;
}

void Kruskal() {
    int cnt = N + M, i, x, y, ans = 0;
    memset(pre, -1, sizeof(int) * (N + M));
    std::sort(E, E + id, cmp);
    for(i = 0; i < id; ++i) {
        if(!same(E[i].u, E[i].v)) {
            unite(E[i].u, E[i].v);
            ans += E[i].w;
            if(--cnt == 1) break;
        }
    }
    printf("%d\n", ans + COST * cnt);
}

int main() {
    // freopen("stdin.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d%d", &N, &M, &R); // G B
        getMap();
        Kruskal();
    }
    return 0;
}

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