POJ 2723 - Get Luffy Out(2-SAT)

#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;
}