poj 2723 Get Luffy Out

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类型：2-sat

题目：http://poj.org/problem?id=2723

来源：Beijing 2005

思路：2-sat判定问题。将一把钥匙看做两个点，不取为i，取为i + 2n。

（1）对于每一对钥匙（u， v），二者最多只能选一个，故有选择v + 2n， u必选，选择u + 2n，v必选。

（2）对于每一个们上的锁（u， v），二者至少选一个，固有选择u，v + 2n必选，选择v，u + 2n必选。

构造好图后，判断解的存在性。

二分答案。

// poj 2723 Get Luffy Out// re ac 484K 47MS#include <iostream>#include <string>#include <queue>#include <stack>#include <algorithm>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>using namespace std;#define FOR(i,a,b) for(i = (a); i < (b); ++i)#define FORE(i,a,b) for(i = (a); i <= (b); ++i)#define FORD(i,a,b) for(i = (a); i > (b); --i)#define FORDE(i,a,b) for(i = (a); i >= (b); --i)#define CLR(a,b) memset(a,b,sizeof(a))#define PB(x) push_back(x)const int N = 5010;const int M = 40010;int low[N], dfn[N];int belong[N];bool inStack[N], vis[N];stack<int> st;int n, m;int step, t;int conflict[N];int du[N];int color[N];int cnt1, cnt2;int head1[N];int door[N][2], key[N][2];int top[N], cnt;struct node {    int s, e;    int len;}c[N];struct edge1 {    int v, nxt;}E1[M];void addedge1(int u, int v) {    E1[cnt1].v = v;    E1[cnt1].nxt = head1[u];    head1[u] = cnt1++;}void tarjan(int u) {    int i;    step++;    st.push(u);    low[u] = dfn[u] = step;    vis[u] = 1;    inStack[u] = 1;    for(i = head1[u]; i != -1; i = E1[i].nxt) {        int x = E1[i].v;        if(!vis[x]) {            tarjan(x);            low[u] = min(low[u], low[x]);        }        else if(inStack[x])            low[u]=min(low[u], dfn[x]);    }    if(low[u] == dfn[u]) {        t++;        while(1) {            int x = st.top();            st.pop();            belong[x] = t;            inStack[x] = 0;            if(x == u)                break;        }    }}void solve() {    int i, j, mid, l = 1, r = m;    int tmp_cnt = cnt1;    while(l <= r) {        mid = (l + r) >> 1;        cnt1 = tmp_cnt;        cnt1 = step = t = 0;        CLR(head1, -1);        CLR(vis, 0);        CLR(inStack, 0);        FORE(i, 1, n) {            addedge1(key[i][0] + 2 * n, key[i][1]);            addedge1(key[i][1] + 2 * n, key[i][0]);        }        FORE(i, 1, mid) {            addedge1(door[i][0], door[i][1] + 2 * n);            addedge1(door[i][1], door[i][0] + 2 * n);        }        while(!st.empty())            st.pop();        FORE(i, 1, 4 * n)            if(!vis[i])                tarjan(i);        bool sign = false;        FORE(i, 1, 2 * n)            if(belong[i] == belong[i + 2 * n]) {                sign = true;                break;            }        if(sign)            r = mid - 1;        else            l = mid + 1;    }    printf("%d\n", l - 1);}int main() {    int u, v, i, j;    while(scanf("%d %d", &n, &m) != EOF, n || m) {        FORE(i, 1, n) {            scanf("%d %d", &key[i][0], &key[i][1]);            ++key[i][0], ++key[i][1];        }        FORE(i, 1, m) {            scanf("%d %d", &door[i][0], &door[i][1]);            ++door[i][0], ++door[i][1];        }        solve();    }    return 0;}