poj 3723 Conscription 最小生成树

Conscription

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8455		Accepted: 2949

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
给你n+m个人，共有r种关系，输入a b c
表示a和b有关系，然后如果其中有一个人先被招募的话，会省下c元钱
每个人初始招募价格为10000
然后问你 最小花费多少钱

题解
直接就连边，然后边的权值为负，然后跑一发最小生成树就好了

代码

int n,m,r;
int x[maxn],y[maxn],d[maxn];
int par[maxn];
int ran[maxn];

void init(int n)
{
    for(int i=0;i<n;i++)
    {
        par[i]=i;
        ran[i]=0;
    }
}
int fin(int x)
{
    if(par[x]==x)
        return x;
    return par[x]=fin(par[x]);
}
void unite(int x,int y)
{
    x=fin(x);
    y=fin(y);
    if(x==y)
        return;
    if(ran[x]<ran[y])
        par[x]=y;
    else
    {
        par[y]=x;
        if(ran[x]==ran[y])
            ran[x]++;
    }
}
bool same(int x,int y)
{
    return fin(x)==fin(y);
}
struct edge
{
    int u,v,cost;
};
bool cmp(edge e1,edge e2)
{
    return e1.cost<e2.cost;
}
edge es[maxn];
int V,E;
int kruskal()
{
    sort(es,es+E,cmp);
    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]};
    }
    cout<<10000*(n+m)+kruskal()<<endl;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&r);
        REP(i,r)
            scanf("%d%d%d",&x[i],&y[i],&d[i]);
        solve();
    }
}