贪心 uvaoj 11134 Fabled Rooks

Problem F: Fabled Rooks

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 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.

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.

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

Output for sample input

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
　　
　　对于一个N*N的棋盘，求放置N个有放置范围的车的一种方案，要求车不能相互攻击。
　　这题有一个很巧妙的性质：行和列是不相关联的，考虑处理两次，贪心即可。
　　
 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 using namespace std;
 6 const int maxn=5010;
 7 
 8 int L[maxn],R[maxn],U[maxn],D[maxn],P[maxn];
 9 int X[maxn],Y[maxn];
10 bool cmp1(int a,int b){
11     if(R[a]!=R[b])return R[a]<R[b];
12     return L[a]<L[b];
13 }
14 
15 bool cmp2(int a,int b){
16     if(D[a]!=D[b])return D[a]<D[b];
17     return U[a]<U[b];
18 }
19 bool vis[maxn];
20 int main(){
21     int n,cnt;
22     while(~scanf("%d",&n)&&n){
23         for(int i=1;i<=n;i++)
24             scanf("%d%d%d%d",&U[i],&L[i],&D[i],&R[i]);
25         for(int i=1;i<=n;i++)P[i]=i;
26         bool OK=true,flag;
27         sort(P+1,P+n+1,cmp1);
28         memset(vis,0,sizeof(vis));cnt=1;
29         for(int i=1,j;i<=n;i++){
30             while(vis[P[cnt]])cnt++;flag=false;
31             for(j=cnt;j<=n;j++)
32                 if(!vis[P[j]]&&i<=R[P[j]]&&i>=L[P[j]]){
33                     flag=true;
34                     break;
35                 }        
36             OK&=flag;
37             if(!OK)break;
38             Y[P[j]]=i;
39             vis[P[j]]=true;
40         }
41         sort(P+1,P+n+1,cmp2);
42         memset(vis,0,sizeof(vis));cnt=1;
43         for(int i=1,j;i<=n;i++){
44             while(vis[P[cnt]])cnt++;flag=false;
45             for(j=cnt;j<=n;j++)
46                 if(!vis[P[j]]&&i<=D[P[j]]&&i>=U[P[j]]){
47                     flag=true;
48                     break;
49                 }        
50             OK&=flag;
51             if(!OK)break;
52             X[P[j]]=i;
53             vis[P[j]]=true;
54         }
55         if(OK){
56             for(int i=1;i<=n;i++)
57                 printf("%d %d\n",X[i],Y[i]);
58             }
59         else
60             printf("IMPOSSIBLE\n");
61     }
62     return 0;
63 }

 
posted @ 2016-03-21 21:02  TenderRun  阅读(150)  评论(0)  编辑  收藏  举报