Poj2723：Get Luffy Out

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;
}