Day5 - D - 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 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

思路：最小生成树板子题，注意这一题是求最大权，取反就是最小生成树，且这一题不是所有点都是连通的，用prim麻烦，直接kruskal最小生成森林即可

const int maxm = 20005;

struct edge {
    int u, v, w;
    edge(int _u=-1, int _v=-1, int _w=0):u(_u), v(_v), w(_w){}
    bool operator<(const edge &a) const {
        return w < a.w;
    }
} Edge[50010];

int fa[maxm], T, N, M, V, R;

void init() {
    for(int i = 0; i <= V; ++i)
        fa[i] = i;
}

int Find(int x) {
    if(fa[x] == x)
        return x;
    return fa[x] = Find(fa[x]);
}

void Union(int x, int y) {
    x = Find(x), y = Find(y);
    if(x != y) fa[x] = y;
}

int main() {
    scanf("%d", &T);
    while(T--) {
        int t1, t2, t3, u, v;
        scanf("%d%d%d", &N, &M, &R);
        V = N + M;
        init();
        int sum = 0;
        for(int i = 0; i < R; ++i) {
            scanf("%d%d%d", &t1, &t2, &t3);
            Edge[i] = edge(t1, t2+N, -t3);
        }
        sort(Edge, Edge+R);
        for(int i = 0; i < R; ++i) {
            u = Edge[i].u, v = Edge[i].v;
            u = Find(u), v = Find(v);
            if(u != v) {
                sum += Edge[i].w;
                Union(u,v);
            }
        }
        printf("%d\n", V*10000 + sum);
    }
    return 0;
}