POJ 2723 Get Luffy Out（2-SAT+二分答案）

Get Luffy Out

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8851		Accepted: 3441

Description

Ratish is a young man who always dreams of being a hero. One day his friend Luffy was caught by Pirate Arlong. Ratish set off at once to Arlong's island. When he got there, he found the secret place where his friend was kept, but he could not go straight in. He saw a large door in front of him and two locks in the door. Beside the large door, he found a strange rock, on which there were some odd words. The sentences were encrypted. But that was easy for Ratish, an amateur cryptographer. After decrypting all the sentences, Ratish knew the following facts:

Behind the large door, there is a nesting prison, which consists of M floors. Each floor except the deepest one has a door leading to the next floor, and there are two locks in each of these doors. Ratish can pass through a door if he opens either of the two locks in it. There are 2N different types of locks in all. The same type of locks may appear in different doors, and a door may have two locks of the same type. There is only one key that can unlock one type of lock, so there are 2N keys for all the 2N types of locks. These 2N keys were divided into N pairs, and once one key in a pair is used, the other key will disappear and never show up again.

Later, Ratish found N pairs of keys under the rock and a piece of paper recording exactly what kinds of locks are in the M doors. But Ratish doesn't know which floor Luffy is held, so he has to open as many doors as possible. Can you help him to choose N keys to open the maximum number of doors?

Input

There are several test cases. Every test case starts with a line containing two positive integers N (1 <= N <= 2¹⁰) and M (1 <= M <= 2¹¹) separated by a space, the first integer represents the number of types of keys and the second integer represents the number of doors. The 2N keys are numbered 0, 1, 2, ..., 2N - 1. Each of the following N lines contains two different integers, which are the numbers of two keys in a pair. After that, each of the following M lines contains two integers, which are the numbers of two keys corresponding to the two locks in a door. You should note that the doors are given in the same order that Ratish will meet. A test case with N = M = 0 ends the input, and should not be processed.

Output

For each test case, output one line containing an integer, which is the maximum number of doors Ratish can open.

Sample Input

3 6
0 3
1 2
4 5
0 1
0 2
4 1
4 2
3 5
2 2
0 0

Sample Output

4

题目链接：POJ 2723

用2-SAT来check的二分答案的一道题目，由于题目的门是从上到下顺序进入的，因此才能二分这个位置，然后检测第1道门~第mid道门是否均能通过。

那么就简单了，首先把钥匙拆成两个点，用和不用，然后一共有2N个钥匙，那么就有4N个点，然后每次建图肯定是先把钥匙的矛盾点建好，然后再用1~mid的门的信息再加入一些边，然后check方案是否存在即可，一开始想不出来怎么做后来发现是按输入的顺序来开门的，然后如果一个钥匙pair为 a与b，如果你选了a，那么b会消失，但a可以一直用，不会消失，感觉以前大一的时候也在某一个网络赛里见过这个题，意思基本相同就题面变一下，可惜当时并不懂2-SAT……写了个暴力dfs结果T了……惨

代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <sstream>
#include <numeric>
#include <cstring>
#include <bitset>
#include <string>
#include <deque>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = (1 << 10) + 10;
const int MAXV = N << 2;
const int MAXE = N * 6;
struct edge
{
	int to, nxt;
	edge() {}
	edge(int _to, int _nxt): to(_to), nxt(_nxt) {}
};
edge E[MAXE];
int head[MAXV], tot;
int vis[MAXV], st[MAXV], top;
int n, m;
int door[N << 1][2], key[N][2];

int rev(int k)
{
	return k < (n << 1) ? k + (n << 1) : k - (n << 1);
}
inline void add(int s, int t)
{
	E[tot] = edge(t, head[s]);
	head[s] = tot++;
}
void init()
{
	CLR(head, -1);
	tot = 0;
	CLR(vis, 0);
}
int dfs(int u)
{
	if (vis[rev(u)])
		return 0;
	if (vis[u])
		return 1;
	vis[u] = 1;
	st[top++] = u;
	for (int i = head[u]; ~i; i = E[i].nxt)
	{
		int v = E[i].to;
		if (!dfs(v))
			return 0;
	}
	return 1;
}
int check()
{
	for (int i = 0; i < (n << 2); ++i)
	{
		top = 0;
		if (!vis[i] && !vis[rev(i)] && !dfs(i))
		{
			while (top)
				vis[st[--top]] = 0;
			if (!dfs(rev(i)))
				return 0;
		}
	}
	return 1;
}
void build(int last)
{
	init();
	for (int i = 0; i < n; ++i) //2n
	{
		add(key[i][0], rev(key[i][1]));
		add(key[i][1], rev(key[i][0]));
	}
	for (int i = 1; i <= last; ++i) //2m
	{
		add(rev(door[i][0]), door[i][1]);
		add(rev(door[i][1]), door[i][0]);
	}
}
int main(void)
{
	int i;
	while (~scanf("%d%d", &n, &m) && (n || m))
	{
		init();
		for (i = 0; i < n; ++i)
			scanf("%d%d", &key[i][0], &key[i][1]);
		for (i = 1; i <= m; ++i)
			scanf("%d%d", &door[i][0], &door[i][1]);
		int L = 0, R = m;
		int ans = 0;
		while (L <= R)
		{
			int mid = MID(L, R);
			build(mid);
			if (check())
			{
				ans = mid;
				L = mid + 1;
			}
			else
				R = mid - 1;
		}
		printf("%d\n", ans);
	}
	return 0;
}