poj3723 Conscription无向图最大权森林问题

1.题目原文

题目链接: http://poj.org/problem?id=3723

Language: Default Conscription Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11499   Accepted: 4070 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 xi, yi and di. There is a blank line before each test case. 1 ≤ N, M ≤ 10000 0 ≤ R ≤ 50,000 0 ≤ xi < N 0 ≤ yi < M 0 < di < 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

2.题目大致意思

需要征募女兵N人，男兵M人。每征募一个人需要花费10000元。但是如果已经征募的人中有一些关系亲密的人，那么可以少花一下钱。给出若干男女之间的1-9999之间的亲密度关系，征募某个人的费用是10000-d(已经征募的人中和自己的亲密度的最大值)。要求通过适当的征募顺序使得征募所有人花费最小。

3.思路与分析

我们设想这样一个无向图:在征募某个人a时，如果使用了a和b之间的关系，就连一条a到b的边。不可能存在一个圈，因为题目要求是男女之间的关系，一个圈最少三个人，三个人不可能满足相邻的人异性，n(n>=3)时类似。因此可知这个图是一片森林(两个连通分量，男生和女生分别在一个连通分量中)。反之，如果给定了一个森林，就可以确定征募的顺序。

因此，把人看成顶点、把关系看成边，这个问题就转化为求解无向图的最大权森林问题。最大权森林问题可以通过把所有边的权值取反之后用最小生成树的算法求解。

4.AC代码

#include<iostream>
#include<cstdio>
#include<set>
#include<algorithm>
using namespace std;
#define maxn 50005
int par[maxn];//父亲
int Rank[maxn];//树的高度
//初始化
void init(int n)
{
    for(int i=0;i<n;i++){
        par[i]=-1;
        Rank[i]=0;
    }
}
//查询父亲，路径压缩
int Find(int x)
{
    int s;
    for(s=x;par[s]!=-1;s=par[s]);
    while(s!=x){
        int temp=par[x];
        par[x]=s;
        x=temp;
    }
    return s;
}
//合并x，y属于的集合
void unite(int x,int y)
{
    x=Find(x);y=Find(y);
    if(x==y) return;
    if(Rank[x]<Rank[y]){
        par[x]=y;
    }
    else{
        par[y]=x;
        if(Rank[x]==Rank[y]) Rank[x]++;
    }
}
bool same(int x,int y)
{
    return Find(x)==Find(y);
}
struct edge
{
    int from;
    int to;
    int cost;
} ;
bool cmp(const edge& e1,const edge& e2)
{
    return e1.cost<e2.cost;
}
int V,E;
edge es[maxn];
int kruscal()
{
    sort(es,es+E,cmp);//按照edge.cost从小到大的顺序进行排序
    init(V);
    int res=0;
    for(int i=0;i<E;i++){
        edge e=es[i];
        if(!same(e.from,e.to)){
            unite(e.from,e.to);
            res+=e.cost;
        }
    }
    return res;
}
int N,M,R;
int x[maxn],y[maxn],d[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&N,&M,&R);
        V=N+M;
        E=R;
        for(int i=0;i<R;i++){
            scanf("%d%d%d",&x[i],&y[i],&d[i]);
            es[i]=(edge){x[i],N+y[i],-d[i]};
        }
        printf("%d\n",10000*(N+M)+kruscal());
    }
    return 0;
}