Mobile Phone Network CodeForces

大意: 无向图, 其中k条边是你的, 边权待定, m条边是你对手的, 边权已知. 求如何设置边权能使最小生成树中, 你的边全被选到, 且你的边的边权和最大. 若有多棵最小生成树优先取你的边.

先将$k$条边合并, 然后按边权从小到大添对手的边, 若连通, 则树链取最小值, 否则合并一下.

正确性其实很显然.

然后对于树链取最小有多种方法, 强制在线可以树剖$O(nlog^2n)$, 可以离线的话可以用倍增$O(nlogn)$, 或者并查集$O(n)$. 

#include <iostream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
const int N = 1e6+10;
int n, k, m, cnt, fa[N];
struct _ {int to,id;} pre[N];
vector<_> g[N];
struct {int u,v,w;} e[N];
int Find(int x) {return fa[x]?fa[x]=Find(fa[x]):x;}
int dep[N], ans[N];

void dfs(int x, int d, int f) {
	dep[x] = d;
	for (_ e:g[x]) if (e.to!=f) { 
		pre[e.to] = {x,e.id};
		dfs(e.to,d+1,x);
	}
}

int main() {
	scanf("%d%d%d", &n, &k, &m);
	int tot = 0;
	while (k--) {
		int u, v;
		scanf("%d%d", &u, &v);
		fa[Find(u)] = Find(v);
		g[u].pb({v,1}),g[v].pb({u,1});
	}
	REP(i,1,m) {
		int u, v, w;
		scanf("%d%d%d", &u, &v, &w);
		int uu = Find(u), vv = Find(v);
		if (uu==vv) e[++cnt]={u,v,w};
		else {
			fa[uu] = vv;
			g[u].pb({v,0}), g[v].pb({u,0});
		}
	}
	dfs(1,0,0);
	memset(fa,0,sizeof fa);
	REP(i,1,cnt) {
		int u = e[i].u, v = e[i].v, w = e[i].w;
		while (Find(u)!=Find(v)) {
			if (dep[u]<dep[v]) swap(u,v);
			if (Find(u)!=Find(pre[u].to)) {
				fa[u] = Find(pre[u].to);
				ans[u] = w;
			}
			u = Find(u);
		}
	}
	ll sum = 0;
	REP(i,1,n) if (pre[i].to&&pre[i].id) { 
		if (!ans[i]) return puts("-1"),0;
		else sum += ans[i];
	}
	printf("%lld
", sum);
}