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[5005],rooky[5005];

int ans[2][5005];
    int n;

int solved (node* x,int p){
    priority_queue<node> q;
    while (!q.empty())
        q.pop();
    for (int i=1;i<=n;i++)
        q.push(x[i]); 
    int cnt=1;
    while (!q.empty() ){
        node a;
        a=q.top();
        q.pop() ;
        if (a.r<cnt)
            return 0;
        if (a.l<cnt){
            a.l=cnt;
            q.push(a);
            continue ;
        } 
        if (a.l>cnt){
            return 0;
        }
        ans[p][a.id]=cnt++;
    }
    return 1;
}

int main (){
    while (cin>>n&&n){
        for (int i=1;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,0)&&solved (rooky,1)){
            for (int i=1;i<=n;i++)
                cout<<ans[0][i]<<" "<<ans[1][i]<<endl;
        }
        else cout<<"IMPOSSIBLE"<<endl;
    }
    return 0;
}