poj 2723【2-SAT+二分】

題意是說，2N把鑰匙，M個門，每個門有兩個鎖，開其中一個就可以，把2N把鑰匙分成Npair，用了其中一把，另一把就不再出現，而用的那把可以迴圈使用，問最多能開多少扇門？

解法：2-SAT+二分

因為開門是有序的，所以二分答案。

設有4N把鎖，其中1~2N表示鎖i（1<=i<=2N）使用，2N+1~4N表示鎖j（1<=j<=2N）拋棄，接著就是激動人心的建圖！先建不能共存的鎖，再對門進行建圖。最後就縮點

#include <vector>#include <list>#include <map>#include <set>#include <queue>#include <string.h>#include <deque>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <limits.h>using namespace std;int lowbit(int t){return t&(-t);}int countbit(int t){return (t==0)?0:(1+countbit(t&(t-1)));}int gcd(int a,int b){return (b==0)?a:gcd(b,a%b);}int max(int a,int b){return a>b?a:b;}int min(int a,int b){return a>b?b:a;}#define LL long long#define PI acos(-1.0)#define N  1200*4#define INF INT_MAX#define eps 1e-8#define FRE freopen("a.txt","r",stdin)int low[N],dfn[N];int belong[N];int key[1200][2],lock[3000][2];bool inStack[N],vis[N];vector<int> v[N];stack<int> st;int n,m;int step;int t;void tarjan(int u){    int i;    step++;    st.push(u);    low[u]=dfn[u]=step;    vis[u]=1;    inStack[u]=1;    for(i=0;i<v[u].size();i++)    {        int x=v[u][i];        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 init(){    int i;    for(i=0;i<=4*n;i++)v[i].clear();    while(!st.empty())st.pop();}bool judge(){    int i,j;    t=0;step=0;    memset(vis,0,sizeof(vis));    memset(inStack,0,sizeof(inStack));    for(i=1;i<=4*n;i++)    if(!vis[i])tarjan(i);    for(i=1;i<=2*n;i++)    if(belong[i]==belong[i+2*n])    return false;    return true;}void build(int mid){    int i,j;    int a,b;    init();    for(i=1;i<=n;i++)    {        a=key[i][1],b=key[i][2];        v[a].push_back(b+2*n);        v[b].push_back(a+2*n);    }    for(i=1;i<=mid;i++)    {        a=lock[i][1],b=lock[i][2];        if(a==b)v[a+2*n].push_back(a);        else        {            v[a+2*n].push_back(b);            v[b+2*n].push_back(a);        }    }}int main(){    while(scanf("%d%d",&n,&m)&&(n+m))    {        int i,j;        for(i=1;i<=n;i++)        {            scanf("%d%d",&key[i][1],&key[i][2]);            key[i][1]++;            key[i][2]++;        }        for(i=1;i<=m;i++)        {            scanf("%d%d",&lock[i][1],&lock[i][2]);            lock[i][1]++;            lock[i][2]++;        }        int l=1,r=m,mid;        int ans;        while(l<=r)        {            mid=(l+r)>>1;            build(mid);            if(judge())            {                ans=mid;                l=mid+1;            }            else            r=mid-1;        }        printf("%d\n",ans);    }    return 0;}