POJ 3723 Conscription

Conscription

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10477		Accepted: 3693

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

题意：须要征募女兵N人，男兵M人。每征募一个人须要花费10000美元，可是假设已经征募的人中有一些关系亲热的人，那么能够少花一些钱。给出若干男女之间亲热关系，征募某个人的费用是10000 - （已经征募的人中和自己亲热度的最大值）。

要求通过适当的征募顺序使得征募全部人所需费用最小。

思路：

把人看做顶点。关系看做边，这个问题就转化为求解无向图中的最大权森林问题。

最大权森林问题能够通过把全部边权取反之后用最小生成树的算法求解。

#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 50000 + 10;

struct edge
{
    int u, v, cost;
   /* edge(int u, int v, int cost){
        this -> u = u;
        this -> v = v;
        this -> cost = cost;
    }*/
};

bool comp(const edge& e1, const edge& e2)
{
    return e1.cost < e2.cost;
}
edge es[maxn];
int N, M, R;
int x[maxn], y[maxn], d[maxn];
int V, E;
int par[maxn];

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

int find(int a)
{
    if (par[a] == a)
        return a;
    else
        return par[a] = find(par[a]);
}

void unite(int a, int b)
{
    a = find(a);
    b = find(b);
    if (a != b)
        par[a] = b;
}

bool same(int a, int b)
{
    return find(a) == find(b);
}

int kruskal()
{
    sort(es, es + E, comp);
    init(V);
    int res = 0;
    for (int i = 0; i < E; i++){
        edge e = es[i];
        if (!same(e.u, e.v)){
            unite(e.u, e.v);
            res += e.cost;
        }
    }
    return res;
}
void solve()
{
    V = N + M;
    E = R;
    for (int i = 0; i < R; i++){
        es[i] = (edge){x[i], N + y[i], -d[i]};
    }
    printf("%d
", 10000 * (N + M) + kruskal());
}
int main()
{
    int t;
    scanf("%d", &t);
    while (t--){
        scanf("%d%d%d", &N, &M, &R);
        for (int i = 0; i < R; i++){
            scanf("%d%d%d", &x[i], &y[i], &d[i]);
        }
        solve();
    }
    return 0;
}