POJ 2723 - Get Luffy Out(2-SAT)

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#题目:

http://poj.org/problem?id=2723

#题意:

2*n把钥匙, 两两配对成n对,用其中一把另一把就会消失.

m个门, 每个门上有两把锁, 打开其中一把就可以打开门. 门打开的顺序应该按照输入的顺序.

求出最多能打开几扇门.

#思路:

每对钥匙只能用其中一把,每个门至少打开一把锁, 符合2-SAT...

1.对于N个点(A,B)表示钥匙,只能用其中的一把, 建边 <A, !B>表示用A不用B, <B,!A>表示用B不用A.

2.对于M个点(A,B)表示锁, 至少打开其中的一把,建边 <!A,B>表示一定打开B, <!B,A>表示一定打开A.

(注意钥匙和锁的连边方式不同)

最后二分求值.

AC.

#include <iostream>#include <cstdio>#include <stack>#include <vector>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 8000;const int MAXM = 8000;struct Key {    int u, v;}key[MAXN];struct Door{    int x, y;}door[MAXN];int n, m;struct edge {    int to, next;}edge[MAXM];int head[MAXN], tot;void init(){    tot = 0;    memset(head, -1, sizeof(head));}void addedge(int u, int v){    edge[tot].to = v;    edge[tot].next = head[u];    head[u] = tot++;}void build(int mmid){    init();    for(int i = 0; i < n/2; ++i) {        addedge(key[i].u, key[i].v+n);        addedge(key[i].v, key[i].u+n);    }    for(int i = 0; i < mmid; ++i) {        addedge(n+door[i].x, door[i].y);        addedge(n+door[i].y, door[i].x);    }}int low[MAXN], dfn[MAXN], belong[MAXN];int scc, Index;stack<int> S;void tarjan(int u){    int v;    low[u] = dfn[u] = ++Index;    S.push(u);    for(int i = head[u]; i != -1; i = edge[i].next) {        v = edge[i].to;        if(!dfn[v]) {            tarjan(v);            if(low[u] > low[v]) low[u] = low[v];        }        else if(!belong[v] && low[u] > dfn[v]) {            low[u] = dfn[v];        }    }    if(low[u] == dfn[u]) {        scc++;        do {            v = S.top(); S.pop();            belong[v] = scc;        }while(v != u);    }}bool solve(int mid){//printf("%d\n", mid);    memset(dfn, 0, sizeof(dfn));    memset(low, 0, sizeof(low));    memset(belong, 0, sizeof(belong));    build(mid);    Index = scc = 0;    for(int i = 0; i < 2*n; ++i) {        if(!dfn[i]) tarjan(i);    }    for(int i = 0; i < n; ++i) {        if(belong[i] == belong[n+i]) {            return false;        }    }    return true;}int main(){//freopen("in", "r", stdin);    while(~scanf("%d %d", &n, &m)) {        if(n == 0 && m == 0) break;        for(int i = 0; i < n; ++i) {            scanf("%d %d", &key[i].u, &key[i].v);        }        for(int i = 0; i < m; ++i) {            scanf("%d %d", &door[i].x, &door[i].y);        }        n = n*2;        int l = 0, r = m;        while(r-l > 1) {            int mid = (l+r)/2;            if(solve(mid)) {                l = mid;            }            else r = mid;        }        if(solve(r)) l = r;                printf("%d\n", l);    }    return 0;}

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