poj 2723 Get Luffy Out 二分+2-SAT-优快云博客

本文介绍了一种结合二分查找与2-可满足性算法(2-SAT)来解决特定类型的组合优化问题的方法。问题背景是有n对钥匙及m扇门，每对钥匙中取一把则另一把消失，每扇门需两把特定钥匙才能打开。通过记录相互排斥的钥匙并使用2-SAT验证可行性，最终确定能按顺序打开的最大门数。

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题目大意：有n对钥匙，每一对拿了一个另外一个就会消失，有m扇门，每一扇门有两个锁，分别对应2*n把钥匙，求按顺序最多可开多少门。
比较明显的模型呢，先把对立边记录下，然后直接二分用2-sat判断，毕竟2-sat不能直接求出答案只能判断是否可行。
二分的时候连前mid扇门就好啦。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
int n, m;
const int N = 3e5 + 5;
unsigned int read()
{
    unsigned int x = 0, f = 1; char ch = getchar();
    while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
    while (ch >= '0'&&ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x*f;
}
int head[N], nxt[N], go[N], val[N];
int oppo[N],low[N],dfn[N];
int a[N], b[N],block[N],stk[N],in[N];
int tot,tim,bl,top;
inline void add(int x, int y)
{
    go[++tot] = y;
    nxt[tot] = head[x];
    head[x] = tot;
}
inline void tarjan(int x)
{
    dfn[x] = low[x] = ++tim;
    stk[++top] = x;
    in[x] = 1;
    for (int i = head[x]; i; i = nxt[i])
    {
        int v = go[i];
        if (!dfn[v])
        {
            tarjan(v);
            low[x] = min(low[v], low[x]);
        }
        else if (in[v]==1) low[x] = min(dfn[v], low[x]);
    }
    if (low[x] == dfn[x])
    {
        bl++;
        int  j=0;
        do
        {
            j = stk[top--];
            in[j] = 0;
            block[j] = bl;
        } while (x != j);
    }
}
inline bool check(int m)
{
    if (!m)return 1;
    tim = tot = 0;
    memset(nxt, 0, sizeof(nxt));
    memset(head, 0, sizeof(head));
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(block, 0, sizeof(block));
    fo(i, 1, m)
    {
        add(oppo[a[i]], b[i]);
        add(oppo[b[i]], a[i]);
    }
    fo(i, 0, 2*n-1)if (!dfn[i])tarjan(i);
    fo(i, 0, 2 * n-1)if (block[i] == block[oppo[i]])return 0;
    return 1;
}
int main()
{
    n = read(), m = read();
    while (n)
    {
        memset(oppo, 0, sizeof(oppo));
        fo(i, 1, n)
        {
            int x = read(), y = read();
            oppo[x] = y;
            oppo[y] = x;
        }
        fo(i, 1, m)a[i] = read(), b[i] = read();
        int l = 0,r = m,mid=(l+r)/2+1;
        while (r!=l)
        {
            if (check(mid)) l = mid ;
            else r = mid - 1; 
            mid = (l + r) / 2+1;
        }
        printf("%d\n", l);
        n = read(), m = read();
    }
    return 0;
}