[WC2016]挑战NPC

Sol

这做法我是想不到\(TAT\)

每个筐子拆成三个相互连边

球向三个筐子连边

然后跑一般图最大匹配

这三个筐子间最多有一个匹配

那么显然每个球一定会放在一个筐子里，一定有一个匹配

如果筐子间有匹配，则有一个半空的筐子，因为它一定只匹配了小于等于\(1\)个球

答案为匹配数\(-n\)

使答案最大即匹配数最大

上带花树就好了

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1005);
const int __(2e5 + 5);
typedef int Arr[_];

IL int Input(){
	RG int x = 0, z = 1; RG char c = getchar();
	for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
	for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
	return x * z;
}

Arr first, match, fa, vis, tim, pre;
int n, m, cnt, idx, ans, E, t1, t2, t3;
queue <int> Q;
struct Edge{
	int to, next;
} edge[__];

IL void Add(RG int u, RG int v){
	edge[cnt] = (Edge){v, first[u]}, first[u] = cnt++;
	edge[cnt] = (Edge){u, first[v]}, first[v] = cnt++;
}

IL int Find(RG int x){
	return x == fa[x] ? x : fa[x] = Find(fa[x]);
}

IL int LCA(RG int x, RG int y){
	++idx, x = Find(x), y = Find(y);
	while(tim[x] != idx){
		tim[x] = idx;
		x = Find(pre[match[x]]);
		if(y) swap(x, y);
	}
	return x;
}

IL void Blossom(RG int x, RG int y, RG int p){
	while(Find(x) != p){
		pre[x] = y, y = match[x];
		if(vis[y] == 2) vis[y] = 1, Q.push(y);
		if(Find(x) == x) fa[x] = p;
		if(Find(y) == y) fa[y] = p;
		x = pre[y];
	}
}

IL int Aug(RG int S){
	for(RG int i = 1; i <= t3; ++i) vis[i] = pre[i] = 0, fa[i] = i;
	while(!Q.empty()) Q.pop();
	Q.push(S), vis[S] = 1;
	while(!Q.empty()){
		RG int u = Q.front(); Q.pop();
		for(RG int e = first[u]; e != -1; e = edge[e].next){
			RG int v = edge[e].to;
			if(Find(v) == Find(u) || vis[v] == 2) continue;
			if(!vis[v]){
				vis[v] = 2, pre[v] = u;
				if(!match[v]){
					for(RG int x = v, lst; x; x = lst)
						lst = match[pre[x]], match[pre[x]] = x, match[x] = pre[x];
					return 1;
				}
				vis[match[v]] = 1, Q.push(match[v]);
			}
			else{
				RG int p = LCA(u, v);
				Blossom(u, v, p);
				Blossom(v, u, p);
			}
		}
	}
	return 0;
}

int main(RG int argc, RG char *argv[]){
	for(RG int T = Input(); T; --T){
		n = Input(), m = Input(), E = Input();
		t1 = n + m, t2 = t1 + m, t3 = t2 + m;
		ans = cnt = idx = 0;
		for(RG int i = 1; i <= t3; ++i) first[i] = -1, match[i] = 0, tim[i] = 0;
		for(RG int i = 1; i <= m; ++i)
			Add(n + i, t1 + i), Add(t1 + i, t2 + i), Add(n + i, t2 + i);
		for(RG int i = 1, u, v; i <= E; ++i){
			u = Input(), v = Input();
			Add(u, n + v), Add(u, t1 + v), Add(u, t2 + v);
		}
		for(RG int i = 1; i <= t3; ++i) if(!match[i]) ans += Aug(i);
		printf("%d\n", ans - n);
	}
	return 0;
}