UVA 11134 Fabled Rooks

贪心+优先队列+问题分解

对x，y 分开处理

当 xl<cnt（当前处理行）时，不能简单的选择cnt，而是应该让xl=cnt 并重新加入优先队列。（y的处理同上）

 #include <iostream>
 #include <cstring>
 #include <algorithm>
 #include <queue>
 using namespace std;

 struct node {
     int l,r;
     int id;
     friend bool operator < (const node &dis,const node &res){
         if (dis.l!=res.l) return dis.l>res.l;
         else return dis.r>res.r;
     }
 }rookx[],rooky[];

 int ans[][];
     int n;

 int solved (node* x,int p){
     priority_queue<node> q;
     while (!q.empty())
         q.pop();
     for (int i=;i<=n;i++)
         q.push(x[i]);
     int cnt=;
     while (!q.empty() ){
         node a;
         a=q.top();
         q.pop() ;
         if (a.r<cnt)
             return ;
         if (a.l<cnt){
             a.l=cnt;
             q.push(a);
             continue ;
         }
         if (a.l>cnt){
             return ;
         }
         ans[p][a.id]=cnt++;
     }
     return ;
 }

 int main (){
     while (cin>>n&&n){
         for (int i=;i<=n;i++){
             cin>>rookx[i].l>>rooky[i].l>>rookx[i].r>>rooky[i].r;
             rookx[i].id=rooky[i].id=i;
         }
         if (solved (rookx,)&&solved (rooky,)){
             for (int i=;i<=n;i++)
                 cout<<ans[][i]<<" "<<ans[][i]<<endl;
         }
         else cout<<"IMPOSSIBLE"<<endl;
     }
     return ;
 }