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 rectangle given by its left-upper corner (xli , yli) and its rightlower 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
1 1 2 2
5 7 8 8
2 2 5 5
2 2 5 5
6 3 8 6
6 3 8 5
6 3 8 8
3 6 7 8
8
1 1 2 2
5 7 8 8
2 2 5 5
2 2 5 5
6 3 8 6
6 3 8 5
6 3 8 8
3 6 7 8
0
Sample Output
1 1
5 8
2 4
4 2
7 3
8 5
6 6
3 7
1 1
5 8
2 4
4 2
7 3
8 5
6 6
3 7
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define N 5010
using namespace std;
struct node {
int xl,xr,yl,yr,x,y,id;
} a[N];
bool vis[N];
//只能按照ri右座標從小到大排序，
//然後從左座標開始選出最小的且未被佔據的座標。
bool cmpx(node x,node y) {
if(x.xr==y.xr)
return x.xl<y.xl;
return x.xr<y.xr;
}
bool cmpy(node x,node y) {
if(x.yr==y.yr)
return x.yl<y.yl;
return x.yr<y.yr;
}
bool cmp(node x,node y) {
return x.id <y.id;
}
int main() {
int n,k;
while(~scanf("%d",&n)&&n) {
for(int i=0; i<n; i++) {
scanf("%d%d%d%d",&a[i].xl,&a[i].yl,&a[i].xr,&a[i].yr);
a[i].x=a[i].y=0;
a[i].id=i;
}
int flag=0;
if(n>1) {
sort(a,a+n,cmpx);
memset(vis,false,sizeof(vis));
for(int i=0; i<n; i++) {
for(k=a[i].xl; k<=a[i].xr; k++) {
if(vis[k]==false) {
a[i].x=k;
vis[k]=true;
break;
}
}
if(a[i].x==0) {
flag=1;
break;
}
}
if(flag==0) {
sort(a,a+n,cmpy);
memset(vis,false,sizeof(vis));
for(int i=0; i<n; i++) {
for(k=a[i].yl; k<=a[i].yr; k++) {
if(vis[k]==false) {
a[i].y=k;
vis[k]=true;
break;
}
}
if(a[i].y==0) {
flag=1;
break;
}
}
}
}
if(n==1) {
a[0].x=a[0].xl;
a[0].y=a[0].yl;
}
if(flag==1)
printf("IMPOSSIBLE\n");
if(flag==0) {
sort(a,a+n,cmp);
for(int i=0; i<n; i++) {
printf("%d %d\n",a[i].x,a[i].y );
}
}
}
}