HDU

题目大意：有N串钥匙，M对锁。每串钥匙仅仅能选择当中一把。怎样选择，才干使开的锁达到最大(锁仅仅能按顺序一对一对开。仅仅要开了当中一个锁就可以)

解题思路：这题跟HDU - 3715 Go Deeper
这题的限制比較简单。都是二选一，2-SAT的裸题，仅仅只是加了二分而已
附上HDU - 3715 Go Deeper题解

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
#define N 4010

struct Pair{
    int x, y;
}P[N];
int lock1[N], lock2[N], S[N];
bool mark[N];
int top, n, m;
vector<int> G[N];

void init() {

    for (int i = 0; i < n; i++) {
        scanf("%d%d", &P[i].x, &P[i].y);
    }
    for (int i = 1; i <= m; i++)
        scanf("%d%d", &lock1[i], &lock2[i]);
}

void AddEdge(int x, int valx, int y, int valy) {
    x = x * 2 + valx;
    y = y * 2 + valy;
    G[x ^ 1].push_back(y);
    G[y ^ 1].push_back(x);
}

bool dfs(int u) {
    if (mark[u ^ 1])
        return false;
    if (mark[u])
        return true;
    mark[u] = true;
    S[++top] = u;
    for (int i = 0; i < G[u].size(); i++)
        if (!dfs(G[u][i]))
            return false;
    return true;
}


bool judge(int mid) {
    for (int i = 0; i < 4 * n; i++)
        G[i].clear();
    for (int i = 0; i < n; i++)
        AddEdge(P[i].x, 0, P[i].y, 0);
    for (int i = 1; i <= mid; i++)
        AddEdge(lock1[i], 1, lock2[i], 1);
    memset(mark, 0, sizeof(mark));

    for (int i = 0; i < 4 * n; i++) {
        if (!mark[i] && !mark[i ^ 1]) {
            top = 0;
            if (!dfs(i)) {
                while (top) mark[S[top--]] = false;
                if (!dfs(i ^ 1))
                    return false;
            }
        }
    }
    return true;
}

void solve() {
    int l = 1, r = m, mid;
    while (l <= r) {
        mid = (l + r) / 2;
        if (judge(mid))
            l = mid + 1;
        else
            r = mid - 1;
    }
    printf("%d
", l - 1);
}
int main() {
    while (scanf("%d%d", &n, &m) != EOF && n + m) {
        init();
        solve();
    }
    return 0;
}