题意

给出 n 对钥匙，每对只能挑一把使用，每把只能用一次，当一对钥匙中的一把被使用后，另一把也就不能再用了；然后给出 m 道门，每个门都有两把钥匙可以打开，问最多能开几道门（按给出的顺序开）。

Sol

这不就是\(HNOI\)超级英雄吗？

上次写的二分图匹配

这次写个\(2-SAT\)

二分答案+\(2-SAT\)判定

注意不要漏

# include <iostream>

# include <stdio.h>

# include <stdlib.h>

# include <string.h>

# include <math.h>

# include <algorithm>

# define RG register

# define IL inline

# define Fill(a, b) memset(a, b, sizeof(a))

using namespace std;

typedef long long ll;

const int _(8050);

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;

}

int n, m, tmp, first[_], cnt, num, x[_], y[_];

int S[_], vis[_], dfn[_], low[_], Index, col[_];

struct Link{

	int u, v;

} link[_];

struct Edge{

	int to, next;

} edge[_ << 1];

IL void Add(RG int u, RG int v){

	edge[cnt] = (Edge){v, first[u]}; first[u] = cnt++;

}

IL void Tarjan(RG int u){

	vis[u] = 1, dfn[u] = low[u] = ++Index, S[++S[0]] = u;

	for(RG int e = first[u]; e != -1; e = edge[e].next){

		RG int v = edge[e].to;

		if(!dfn[v]) Tarjan(v), low[u] = min(low[u], low[v]);

		else if(vis[v]) low[u] = min(low[u], dfn[v]);

	}

	if(dfn[u] != low[u]) return;

	RG int v = S[S[0]--]; col[v] = ++num, vis[v] = 0;

	while(v != u) v = S[S[0]--], col[v] = num, vis[v] = 0;

}

IL int Check(RG int mid){

	Fill(first, -1), Fill(dfn, 0), Fill(col, 0), cnt = num = 0;

	for(RG int i = 0; i < n; ++i) Add(x[i], y[i] + tmp), Add(y[i], x[i] + tmp);

	for(RG int i = 1; i <= mid; ++i)

		Add(link[i].u + tmp, link[i].v), Add(link[i].v + tmp, link[i].u);

	for(RG int i = 0, t = tmp << 1; i < t; ++i) if(!dfn[i]) Tarjan(i);

	for(RG int i = 0; i < tmp; ++i) if(col[i] == col[i + tmp]) return 0;

	return 1;

}

int main(RG int argc, RG char* argv[]){

	while(233){

		n = Input(), m = Input();

		if(!(n + m)) break;

		tmp = n << 1;

		for(RG int i = 0; i < n; ++i) x[i] = Input(), y[i] = Input();

		for(RG int i = 1; i <= m; ++i) link[i] = (Link){Input(), Input()};

		RG int l = 0, r = m, ans = 0;

		while(l <= r){

			RG int mid = (l + r) >> 1;

			if(Check(mid)) ans = mid, l = mid + 1;

			else r = mid - 1;

		}

		printf("%d\n", ans);

	}

	return 0;

}