POJ 3723 Conscription 克鲁斯卡尔+并查集

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Conscription

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11318 Accepted: 3997

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

25 5 84 3 68311 3 45830 0 65920 1 30633 3 49751 3 20494 2 21042 2 7815 5 102 4 98203 2 62363 1 88642 4 83262 0 51562 0 14634 1 24390 4 43733 4 88892 4 3133

Sample Output

7107154223

Source

POJ Monthly Contest – 2009.04.05, windy7926778

题意：招人，招收每个人需要花费10000元，男女生之间存在关系的话，则可以少花关系值的钱。关系不可重复使用，问最终所花的最少的招人费用。

分析：要使招人费用最少，就要使得关系和最大，这里将每个关系抽象为一条边，最终根据这些边生成的子集树的代价最大即为所求最大。这里使用克鲁斯卡尔算法，克鲁斯卡尔算法求最小代价生成树是将边排序，每次选出最小的边判断其端点是否是属于两个不同的子树，是的话则加入此边，不是的会就会造成回路，此边舍弃。此题所求为最大代价生成树，原理相同，将边以非递减顺序排序即可。判断是否属于同一子树，在此用并查集实现即可。注意男女序号都是从0开始，并查集实现时需要区分男女序号。见AC代码：

//该题的意思是要最终所花费用最小 即减少的费用最大//即求最大生成树  两个人之间的关系抵消掉的值 即为边值//要让边值的和最大//按照边长  从大到小排序//每次选取一条边  该边的一点已被选上  另一点未被选上//使用的是克鲁斯卡尔算法算法#include<stdio.h>#include<algorithm>using namespace std;const int maxn=50005;int n,m,r,maxtree,res;//res表示最大生成树的边长之和int pre[maxn];struct  edge{int u,v,w;} a[maxn];int cmp(edge a ,edge b){return a.w>b.w;}int find(int x)                                      {int r=x;while (pre[r]!=r)r=pre[r];int i=x,j ;while( i != r ){j=pre[i];pre[i]=r ;i=j;}return r ;}void join(int x,int y){int fx=find(x),fy=find(y);if(fx!=fy)pre[fx]=fy;}void kruskal(){maxtree = 0;for(int i = 0; i < r; i++){int fa=find(a[i].u);int fb=find(a[i].v+n);if(fa==fb)continue;else{join(a[i].u,a[i].v+n);//printf("加上%d\n",a[i].w);maxtree+=a[i].w;}}}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",&a[i].u,&a[i].v,&a[i].w);pre[a[i].u]=a[i].u;pre[a[i].v+n]=n+a[i].v;}sort(a, a + r, cmp);kruskal();int res=(n+m)*10000-maxtree;printf("%d\n",res);}}

特记下，以备后日回顾。

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