UVA - 11134 Fabled Rooks[贪心 问题分解]

We would like to place n rooks, 1 ≤ n ≤ 5000, on a n × n board subject to the following restrictions

• The i-th rook can only be placed within the rectan- gle given by its left-upper corner (xli,yli) and its right- lower corner (xri,yri), where 1 ≤ i ≤ n, 1 ≤ xli ≤ xri ≤n,1≤yli ≤yri ≤n.
• No two rooks can attack each other, that is no two rooks can occupy the same column or the same row.

  Input

  The input consists of several test cases. The first line of each
  of them contains one integer number, n, the side of the board. n lines follow giving the rectangles
  where the rooks can be placed as described above. The i-th line among them gives xli, yli, xri, and
  yri. The input file is terminated with the integer ‘0’ on a line by itself.

  Output

  Your task is to find such a placing of rooks that the above conditions are satisfied and then output n
  lines each giving the position of a rook in order in which their rectangles appeared in the input. If there
  are multiple solutions, any one will do. Output ‘IMPOSSIBLE’ if there is no such placing of the rooks.

  Sample Input

  8
  1122
  5788
  2255
  2255
  6386
  6385
  6388
  3678
  8
  1122
  5788
  2255
  2255
  6386
  6385
  6388
  3678
  0

  Sample Output

  11
  58
  24
  42
  73
  85
  66
  37
  11
  58
  24
  42
  73
  85
  66
  37

在n*n的棋盘上面放n个车,能否使他们互相不攻击(即不能在同一行一列),并且第i个车必须落在第i的矩形范围(xl,yl, xr,yr)之内

行列可以分开求解，变成了区间问题

一开始想按r小大排序，r相同按l排序，然后依次选择行了

然而，应该是“对于每个位置，选择合法且r最小的”，这样排序并不能保证

只好n^2暴力找了

//
//  main.cpp
//  uva11134
//
//  Created by Candy on 10/17/16.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N=5e3+5,INF=1e9;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();}
    return x*f;
}
int n;
int x[N],y[N];
struct data{
    int id;
    int l1,r1,l2,r2;
}a[N];
//bool cmp1(data &x,data &y){
//    if(x.l1==y.l1) return x.r1<y.r1;
//    else return x.l1<y.l1;
//}
//bool cmp2(data &x,data &y){
//    if(x.l2==y.l2) return x.r2<y.r2;
//    return x.l2<y.l2;
//}
//bool solve(){
//    sort(a+1,a+1+n,cmp1);
//    int p=1;
//    for(int i=1;i<=n;i++){
//        //printf("hi%d %d %d %d\n",i,a[i].l1,a[i].r1,a[i].id);
//        if(a[p].l1<=i&&i<=a[p].r1) x[a[p++].id]=i;
//        else if(a[p].r1<i||a[p].l1>i) return 0;
//    }
//
//    sort(a+1,a+1+n,cmp2);
//    p=1;
//    for(int i=1;i<=n;i++){
//        //printf("ih%d %d %d %d\n",i,a[i].l2,a[i].r2,a[i].id);
//        if(a[p].l2<=i&&i<=a[p].r2) y[a[p++].id]=i;
//        else if(a[p].r2<i||a[p].l2>i) return 0;
//    }
//    return 1;
//}
bool sol(){
    memset(x,0,sizeof(x));
    memset(y,0,sizeof(y));
    for(int i=1;i<=n;i++){
        int rook=0,mnr=INF;
        for(int j=1;j<=n;j++)
            if(!x[j]&&a[j].l1<=i&&a[j].r1<mnr) rook=j,mnr=a[j].r1;
        //printf("rook1 %d\n",rook);
        if(rook==0||a[rook].r1<i) return false;
        x[rook]=i;
    }
    for(int i=1;i<=n;i++){
        int rook=0,mnr=INF;
        for(int j=1;j<=n;j++)
            if(!y[j]&&a[j].l2<=i&&a[j].r2<mnr) rook=j,mnr=a[j].r2;
        //printf("rook2 %d\n",rook);
        if(rook==0||a[rook].r2<i) return false;
        y[rook]=i;
    }
    return 1;
}
int main(int argc, const char * argv[]) {
    while((n=read())){
        for(int i=1;i<=n;i++){
            a[i].id=i;
            a[i].l1=read();a[i].l2=read();a[i].r1=read();a[i].r2=read();
        }
        if(sol()){
            for(int i=1;i<=n;i++) printf("%d %d\n",x[i],y[i]);
        }else printf("IMPOSSIBLE\n");
    }

    return 0;
}