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）
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 #include <queue> 5 using namespace std; 6 7 struct node { 8 int l,r; 9 int id; 10 friend bool operator < (const node &dis,const node &res){ 11 if (dis.l!=res.l) return dis.l>res.l; 12 else return dis.r>res.r; 13 } 14 }rookx[5005],rooky[5005]; 15 16 int ans[2][5005]; 17 int n; 18 19 int solved (node* x,int p){ 20 priority_queue<node> q; 21 while (!q.empty()) 22 q.pop(); 23 for (int i=1;i<=n;i++) 24 q.push(x[i]); 25 int cnt=1; 26 while (!q.empty() ){ 27 node a; 28 a=q.top(); 29 q.pop() ; 30 if (a.r<cnt) 31 return 0; 32 if (a.l<cnt){ 33 a.l=cnt; 34 q.push(a); 35 continue ; 36 } 37 if (a.l>cnt){ 38 return 0; 39 } 40 ans[p][a.id]=cnt++; 41 } 42 return 1; 43 } 44 45 int main (){ 46 while (cin>>n&&n){ 47 for (int i=1;i<=n;i++){ 48 cin>>rookx[i].l>>rooky[i].l>>rookx[i].r>>rooky[i].r; 49 rookx[i].id=rooky[i].id=i; 50 } 51 if (solved (rookx,0)&&solved (rooky,1)){ 52 for (int i=1;i<=n;i++) 53 cout<<ans[0][i]<<" "<<ans[1][i]<<endl; 54 } 55 else cout<<"IMPOSSIBLE"<<endl; 56 } 57 return 0; 58 }