URAL-1997 Those are not the droids you're looking for 二分匹配

　　题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1997

　　题意：记录了n个人进出门的时间点，每个人在房子里面待的时间要么小于等于a，要么大于等于b，询问能否对进出门的时间点找到一个合适的匹配。

　　对于满足的要求建立边即可，然后看是否能找到最大匹配。

 //STATUS:C++_AC_109MS_4361KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 int vis[N],y[N],w[N][N],t[N][],id[N],l[N],r[N];
 int n,a,b,cnt1,cnt2;

 int dfs(int u)
 {
     int v;
     for(v=;v<cnt1;v++){
         if(vis[v] || !w[u][v])continue;
         vis[v]=;
         if(y[v]==- || dfs(y[v])){
             y[v]=u;
             return ;
         }
     }
     return ;
 }

 int main()
 {
  //   freopen("in.txt","r",stdin);
     int i,j,L,R,ok;
     while(~scanf("%d%d",&a,&b))
     {
         L=b+,R=a-;
         scanf("%d",&n);
         cnt1=cnt2=;
         for(i=;i<n;i++){
             scanf("%d%d",&t[i][],&t[i][]);
             if(t[i][]){
                 l[cnt1]=i;
                 id[i]=cnt1++;
             }
             else {
                 r[cnt2]=i;
                 id[i]=cnt2++;
             }
         }
         mem(w,);
         for(i=;i<n;i++){
             if(t[i][]==)continue;
             for(j=i-;j>=;j--){
                 if(t[j][]==)continue;
                 if(t[i][]-t[j][]>=L && t[i][]-t[j][]<=R)continue;
                 w[id[i]][id[j]]=;
             }
         }
         mem(y,-);ok=;
         for(i=;i<cnt1;i++){
             mem(vis,);
             if(dfs(i)==){
                 ok=;
                 break;
             }
         }

         if(ok){
             printf("No reason\n");
             for(i=;i<cnt1;i++){
                 printf("%d %d\n",t[r[i]][],t[ l[ y[i] ] ][]);
             }
         }
         else printf("Liar\n");

     }
     return ;
 }