图论补完计划poj 2723（2-SAT）

Posted 2020-09-03 hymscoty

Get Luffy Out

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8688		Accepted: 3371

Description

Ratish is a young man who always dreams of being a hero. One day his friend Luffy was caught by Pirate Arlong. Ratish set off at once to Arlong‘s island. When he got there, he found the secret place where his friend was kept, but he could not go straight in. He saw a large door in front of him and two locks in the door. Beside the large door, he found a strange rock, on which there were some odd words. The sentences were encrypted. But that was easy for Ratish, an amateur cryptographer. After decrypting all the sentences, Ratish knew the following facts:

Behind the large door, there is a nesting prison, which consists of M floors. Each floor except the deepest one has a door leading to the next floor, and there are two locks in each of these doors. Ratish can pass through a door if he opens either of the two locks in it. There are 2N different types of locks in all. The same type of locks may appear in different doors, and a door may have two locks of the same type. There is only one key that can unlock one type of lock, so there are 2N keys for all the 2N types of locks. These 2N keys were divided into N pairs, and once one key in a pair is used, the other key will disappear and never show up again.

Later, Ratish found N pairs of keys under the rock and a piece of paper recording exactly what kinds of locks are in the M doors. But Ratish doesn‘t know which floor Luffy is held, so he has to open as many doors as possible. Can you help him to choose N keys to open the maximum number of doors?

Input

There are several test cases. Every test case starts with a line containing two positive integers N (1 <= N <= 2¹⁰) and M (1 <= M <= 2¹¹) separated by a space, the first integer represents the number of types of keys and the second integer represents the number of doors. The 2N keys are numbered 0, 1, 2, ..., 2N - 1. Each of the following N lines contains two different integers, which are the numbers of two keys in a pair. After that, each of the following M lines contains two integers, which are the numbers of two keys corresponding to the two locks in a door. You should note that the doors are given in the same order that Ratish will meet. A test case with N = M = 0 ends the input, and should not be processed.

Output

For each test case, output one line containing an integer, which is the maximum number of doors Ratish can open.

Sample Input

3 6
0 3
1 2
4 5
0 1
0 2
4 1
4 2
3 5
2 2
0 0

Sample Output

4

题意
现在有N对钥匙（一共2N把，编号从0到2N-1），每对钥匙只能选其中一把。另有M扇门，每扇门能够被两把钥匙a，b中的任意一把打开，问能开的最大门数是多少（开门为递增顺序）

思路
2-SAT,找到矛盾关系。①如果a，b出现在同一对钥匙里，只能拿a或拿b。②如果c，d能打开同一扇门，为了最大化开门数，我们假定，不用c开门或不用d开门。
蕴含式表示形式：a->!b和b->!a，!c->d和!d->c。按照这个建图，求scc，如果i和i+2*n在一个强连通分量里（即i和!i在同一个主合取范式里）则无解。
由于求的是最大值，所以二分处理。

  1 #include <iostream>
  2 #include <vector>
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <queue>
  6 #include <map>
  7 
  8 using namespace std;
  9 
 10 typedef pair<int,int> P;
 11 
 12 const int maxn=6000;
 13 
 14 vector<int> G[maxn];
 15 vector<int> rG[maxn];
 16 vector<int> vs;
 17 
 18 int n,m;
 19 
 20 bool used[maxn];
 21 int cmp[maxn];
 22 
 23 inline int read(){
 24     int x=0,f=1;char ch=getchar();
 25     while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)f=-1;ch=getchar();}
 26     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
 27     return x*f;
 28 }
 29 
 30 void init(){
 31     for(int i=0;i<maxn;i++){
 32         G[i].clear();
 33         rG[i].clear();
 34         cmp[i]=0;
 35     }
 36 }
 37 
 38 void add_edge(int s,int t){
 39     G[s].push_back(t);
 40     rG[t].push_back(s);
 41 }
 42 
 43 void dfs(int s){
 44     used[s]=true;
 45     for(int i=0;i<G[s].size();i++){
 46         if(!used[G[s][i]]) dfs(G[s][i]);
 47     }
 48     vs.push_back(s);
 49 }
 50 
 51 void rdfs(int s,int k){
 52     used[s]=true;
 53     cmp[s]=k;
 54     for(int i=0;i<rG[s].size();i++){
 55         if(!used[rG[s][i]]) rdfs(rG[s][i],k);
 56     }
 57 }
 58 
 59 int scc(){
 60     vs.clear();
 61     memset(used,0,sizeof(used));
 62     for(int i=0;i<4*n;i++){
 63         if(!used[i]) dfs(i);
 64     }
 65     memset(used,0,sizeof(used));
 66     int k=0;
 67     for(int i=vs.size()-1;i>=0;i--){
 68         if(!used[vs[i]]) rdfs(vs[i],++k);
 69     }
 70     return k;
 71 }
 72 
 73 P key[maxn];
 74 P door[maxn];
 75 
 76 bool solve(int num){
 77     init();
 78     for(int i=0;i<n;i++){
 79         add_edge(key[i].first,key[i].second+2*n);
 80         add_edge(key[i].second,key[i].first+2*n);
 81     }
 82     for(int i=0;i<num;i++){
 83         add_edge(door[i].first+2*n,door[i].second);
 84         add_edge(door[i].second+2*n,door[i].first);
 85     }
 86     int k=scc();
 87     for(int i=0;i<2*n;i++){
 88         if(cmp[i]==cmp[i+2*n]){
 89             return false;
 90         }
 91     }
 92     return true;
 93 }
 94 
 95 int main(){
 96     while(true){
 97         n=read();m=read();
 98         if(n==0&&m==0) break;
 99         for(int i=0;i<n;i++){
100             key[i].first=read();
101             key[i].second=read();
102         }
103         for(int i=0;i<m;i++){
104             door[i].first=read();
105             door[i].second=read();
106         }
107         int l=0,r=m;
108         int ans=0;
109         while(l<=r){
110             int mid=(l+r)/2;
111             if(solve(mid)){
112                 ans=max(ans,mid);
113                 l=mid+1;
114             }
115             else r=mid-1;
116         }
117         printf("%d\n",ans);
118     }
119     return 0;
120 }