Get Luffy Out（2 - sat + 二分）

Get Luffy Out

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7299		Accepted: 2781

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

 

       题意：

       每个 case 给出 N （1 ~ 2 ^ 10）和 M （1 ~ 2 ^ 11）。代表有 N 组钥匙配对关系和 M 道门，每道门都有两把可开钥匙，选其中一把开即可通过这道门。每对配对钥匙间只能用其中一条，输出最大能通过几道门。直到输出 0 0结束。

 

       思路：

       2 - sat。一个个从头到尾 2 - sat 的话无疑会TLE，故要二分答案，每二分一次就要重新建图一次。记得要 clear 可变数组。节点数是 2 * n 个，故 clear 的时候要 clear 2 * 2 * N = 4 * N 个。

 

       AC：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <stack>

using namespace std;

const int NMAX = 5000;

int n, m;
int door[NMAX][4], d[NMAX][4];
vector<int> v[NMAX];

int dfs_clock, scc_cnt;
int pre[NMAX], low[NMAX], cmp[NMAX];
stack<int> s;

void dfs(int u) {
        pre[u] = low[u] = ++dfs_clock;
        s.push(u);

        for (int i = 0; i < v[u].size(); ++i) {
                if (!pre[ v[u][i] ]) {
                        dfs( v[u][i] );
                        low[u] = min(low[u], low[ v[u][i] ]);
                } else if (!cmp[ v[u][i] ]) {
                        low[u] = min(low[u], pre[ v[u][i] ]);
                }
        }

        if (pre[u] == low[u]) {
                ++scc_cnt;
                for (;;) {
                        int x = s.top(); s.pop();
                        cmp[x] = scc_cnt;
                        if (x == u) break;
                }
        }
}

void scc() {
        dfs_clock = scc_cnt = 0;
        memset(pre, 0, sizeof(pre));
        memset(cmp, 0, sizeof(cmp));

        for (int i = 0;i < 4 * n; ++i) {  //是4 * N
                if (!pre[i]) dfs(i);
        }
}

bool C(int k) {

        for (int i = 0; i < 4 * n; ++i)  //是4 * N
                v[i].clear();

        for (int i = 1; i <= n; ++i) {
                v[ d[i][1] ].push_back( d[i][2] + 2 * n);
                v[ d[i][2] ].push_back( d[i][1] + 2 * n);
        }

        for (int i = 1; i <= k; ++i) {
                v[ door[i][1] + 2 * n].push_back( door[i][2] );
                v[ door[i][2] + 2 * n].push_back( door[i][1] );
        }

        scc();

        for (int i = 0; i < 2 * n; ++i) {
                if (cmp[i] == cmp[i + 2 * n]) return false;
        }

        return true;
}

int main() {

        while (~scanf("%d%d", &n, &m) && (n + m)) {

                for (int i = 1; i <= n; ++i)
                        scanf("%d%d", &d[i][1], &d[i][2]);

                for (int i = 1; i <= m; ++i)
                        scanf("%d%d", &door[i][1], &door[i][2]);

                int l = 0, r = m + 1;
                while (r - l > 1) {
                        int mid = l + (r - l) / 2;
                        if (C(mid)) l = mid;
                        else r = mid;
                }

                printf("%d\n", l);
        }

        return 0;
}