POJ 3723 Conscription——最小生成树

边权取反，跑一遍MST，然后加上10000*（n+m），注意编号问题

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 50005;
struct Edge {
int u, v, cost;
bool operator < (const Edge &e) const {
return cost < e.cost;
}
}edges[maxn];
int T, n, m, r;
int fa[maxn];
void init() {
for (int i = 0; i <= n + m; i++) fa[i] = i;
}
int query(int x) { return fa[x] == x ? x : fa[x] = query(fa[x]); }

int main() {
scanf("%d", &T);
while (T--) {
scanf("%d %d %d", &n, &m, &r);
init();
for (int i = 1; i <= r; i++) {
scanf("%d %d %d", &edges[i].u, &edges[i].v, &edges[i].cost);
edges[i].v += n;
edges[i].cost = -edges[i].cost;
}
sort(edges + 1, edges + r + 1);
int ans = 0;
for (int i = 1; i <= r; i++) {
int u = edges[i].u, v = edges[i].v, cost = edges[i].cost;
u = query(u), v = query(v);
if (u == v) continue;
fa[u] = v;
ans += cost;
}
printf("%d\n", ans + 10000 * (n + m));
}
return 0;
}