poj 3723 最大权森林问题
2017-09-06 17:49
381 查看
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
可以把人看作顶点, 关系看作边,这个问题就可以转化为求解无向图中的最大权森林问题.最大权森林问题可以通过把所有边取反之后用最小生成树的算法求解.
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
题解:
在征募某个人a时,如果使用来a和b之间的关系,那么就连一条a到b的边.假设这个图中存在圈,那么无论以什么顺序征募这个圈上的所有人, 都会产生矛盾.(只有男女关系是产生不了圈的…)因此可以直到这个图是一片森林. 反之,如果给了一片森林那么就可以使用对应的关系确定征募的顺序.可以把人看作顶点, 关系看作边,这个问题就可以转化为求解无向图中的最大权森林问题.最大权森林问题可以通过把所有边取反之后用最小生成树的算法求解.
代码:
#include <iostream> #include <algorithm> #include <cstdio> using namespace std; const int MAX_R = 6e5; struct edge { int u,v,cost; }; bool cmp(const edge& e1,const edge& e2) { return e1.cost<e2.cost; } edge es[MAX_R]; int V,E; //顶点数和边数 int N,M,R; int x[MAX_R],y[MAX_R],d[MAX_R]; int par[MAX_R],rank[MAX_R]; void init_union_set(int v) { for(int i=0;i<v;i++){ par[i]=i; rank[i]=0; } } int find(int x){ if(par[x]==x)return x; else return par[x]=find(par[x]); } void union_set(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); } int kruskal() { sort(es,es+E,cmp); init_union_set(V); int res=0; for(int i=0;i<E;i++) { edge e = es[i]; if(!same(e.u,e.v)) { union_set(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].u=x[i]; es[i].v=N+y[i]; es[i].cost=-d[i]; } printf("%d\n",10000*(N+M)+kruskal()); } int main() { int T; cin>>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; }
相关文章推荐
- POJ3723 Conscription , 最大权森林问题 ->最小生成树问题
- POJ 3723-Constription [最大权森林] 《挑战程序设计竞赛》2.5
- POJ 3723 Conscription(最大生成森林)
- POJ - 3723 Conscription(kruskal+最大权森林)
- POJ 3723 Conscription 【最大生成树||最大权森林】
- poj - 3723 Conscription(最大权森林)
- POJ 3723 Conscription (最大权森林 + Kruskal算法)
- Poj 3723 Conscription -- 最大生成树(森林)
- poj 3723(最大权森林)
- 最大权森林·POJ-3723·Conscription
- poj 3723 最大权森林 最小生成树的运用
- poj 3723 Conscription 【最大生成树|最大权森林】
- POJ 1050 To the Max(dp 最大子矩阵和/最大子段和问题)
- POJ 3692 Kindergarten (补图是二分图的最大团问题)
- POJ_2593最大两不想交子段和问题
- 最大匹配问题 POJ 1274
- POJ1336 The K-League[最大流 公平分配问题]
- 2014湘潭全国邀请赛I题 Intervals /POJ 3680 / 在限制次数下取有权区间使权最大/小问题(费用流)
- POJ 1050 二维最大连续子向量问题
- poj 3723 Conscription(最大生成树)