POJ 3723 Conscription （kruskal最大生成森林）
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8479 | Accepted: 2959 |
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
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
求招募亲密度之间最大的
- - prim不能求森林 毕竟点贪心
然后经典技巧 边取负 即可求最大权森林
AC代码如下：
//
// POJ 3723 Conscription
//
// Created by TaoSama on 2015-03-20
// Copyright (c) 2015 TaoSama. All rights reserved.
//
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
#define CLR(x,y) memset(x, y, sizeof(x))
using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;
int n, m, r;
struct Edge {
int u, v, cost;
bool operator<(const Edge& rhs) const {
return cost < rhs.cost;
}
} G[50005];
int par[20005], rank[20005];
void init(int n) {
for(int i = 0; i < n; ++i) {
par[i] = i;
rank[i] = 0;
}
}
int find(int x) {
if(par[x] == x) return x;
return par[x] = find(par[x]);
}
void unite(int x, int y) {
x = find(x); y = find(y);
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() {
int ret = 0;
sort(G + 1, G + 1 + r);
init(n + m);
for(int i = 1; i <= r; ++i) {
Edge &e = G[i];
if(!same(e.u, e.v)) {
ret += e.cost;
unite(e.u, e.v);
}
}
return ret;
}
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
int t; scanf("%d", &t);
while(t--) {
scanf("%d%d%d", &n, &m, &r);
for(int i = 1; i <= r; ++i) {
int x, y, v; scanf("%d%d%d", &x, &y, &v);
G[i] = (Edge) {x, y + n, -v};
}
int ans = 10000 * (n + m) + kruskal();
printf("%d\n", ans);
}
return 0;
}