HDU - 1816 Get Luffy Out *(二分 ＋ 2-SAT)

题目大意：有N串钥匙，M对锁。每串钥匙只能选择其中一把，如何选择，才能使开的锁达到最大(锁只能按顺序一对一对开，只要开了其中一个锁即可)

解题思路：这题跟HDU - 3715 Go Deeper
这题的限制比较简单，都是二选一，2-SAT的裸题，只不过加了二分而已
附上HDU - 3715 Go Deeper题解

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;#define N 4010struct 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\n", l - 1);}int main() {    while (scanf("%d%d", &n, &m) != EOF && n + m) {        init();        solve();    }    return 0;}