【算法练习】POJ - 2723 Get Luffy Out （2-SAT）

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博客围绕一道钥匙相关题目展开，包含题意、题解和代码部分。题解中指出题目看似不难，但存在一个坑点，即题目提到有nn对钥匙，对应2∗n2∗n个钥匙和4∗n4∗n个布尔变量，博主曾被此坑。

题意

题解

题没什么难的。说一下一个坑点。
题目说的是有 $n$ 对钥匙，就有 $2*n$ 个钥匙，就有 $4*n$ 个布尔变量。
被这个坑了半天。

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
const int nmax = 1e5+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const ull p = 67;
const ull MOD = 1610612741;
int head[nmax], tot;
struct edge {
    int to, nxt;
}e[nmax];
typedef pair<int, int> pii;
int low[10005], dfn[10005], color[10005], ss[10005], st, dfs_clock, scc, doorcnt, keyscnt;
bool instack[10005];
pii doors[10005], keys[10005];
int n , m;
void add_edge(int u, int v) {
    e[tot].to = v;
    e[tot].nxt = head[u];
    head[u] = tot++;
}
void init() {
    memset(head ,-1, sizeof head);
    memset(low, 0, sizeof low);
    memset(dfn, 0, sizeof dfn);
    memset(color, 0, sizeof color);
    memset(head, -1, sizeof head);
    tot = st = dfs_clock = scc = 0;
}
void tarjan(int u) {
    low[u] = dfn[u] = ++dfs_clock;
    ss[st++] = u ;
    instack[u] = true;
    for(int i = head[u]; i != -1; i = e[i].nxt) {
        int v = e[i].to;
        if(!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        } else if(instack[v]) {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if(low[u] == dfn[u]) {
        scc ++;
        int tmp = 0;
        while(1) {
            tmp = ss[--st];
            color[tmp] = scc;
            instack[tmp] = false;
            if(tmp == u)
                break;
        }
    }
}
bool check(int x) {
    memset(low, 0, sizeof low);
    memset(dfn, 0, sizeof dfn);
    memset(instack, 0, sizeof instack);
    memset(color, 0, sizeof color);
    memset(head, -1, sizeof head);
    tot = st = dfs_clock = scc = 0;
    for(int i = 1; i <= n; ++i) {
        add_edge(keys[i].first, keys[i].second + 2 * n);
        add_edge(keys[i].second, keys[i].first + 2 * n);
    }
    for(int i = 1; i <= x; ++i) {
        add_edge(doors[i].first + 2 * n, doors[i].second);
        add_edge(doors[i].second + 2 * n, doors[i].first);
    }
    for(int i = 1; i <= 4 * n; ++i) {
        if(!dfn[i])
            tarjan(i);
    }
    for(int i = 1; i <= 2 * n; ++i) {
        if(color[i] == color[i+ 2 * n])
            return false;
    }
    return true;
}
int main(){
    while(scanf("%d %d", &n, &m) != EOF) {
        if(n == 0 && m == 0)
            break;
        init();
        int u , v;
        for(int i = 1; i <= n; ++i) {
            scanf("%d %d", &keys[i].first, &keys[i].second);
            keys[i].first ++;
            keys[i].second ++;
        }

        for(int i = 1; i <= m; ++i) {
            scanf("%d %d", &doors[i].first, &doors[i].second);
            doors[i].first ++;
            doors[i].second ++;
        }
        int ans = 0, l = 0 , r = m;
        while(l <= r) {
            int mid = (l + r) >>1;
            if(check(mid)) {
                ans = mid;
                l = mid + 1;
            }else r = mid - 1;
        }
        printf("%d\n",ans);
    }
    return 0;
}