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ZOJ 3209 Treasure Map （Dancing Links）

2024-07-06 04:18:50   225人阅读

Treasure Map
Time Limit: 2 Seconds      Memory Limit: 32768 KB

Your boss once had got many copies of a treasure map. Unfortunately, all the copies are now broken to many rectangular pieces, and what make it worse, he has lost some of the pieces. Luckily, it is possible to figure out the position of each piece in the original map. Now the boss asks you, the talent programmer, to make a complete treasure map with these pieces. You need to make only one complete map and it is not necessary to use all the pieces. But remember, pieces are not allowed to overlap with each other (See sample 2).

Input

The first line of the input contains an integer T (T <= 500), indicating the number of cases.

For each case, the first line contains three integers n m p (1 <= n, m <= 30, 1 <= p <= 500), the width and the height of the map, and the number of pieces. Then p lines follow, each consists of four integers x1 y1 x2 y2 (0 <= x1 < x2 <= n, 0 <= y1 < y2 <= m), where (x1, y1) is the coordinate of the lower-left corner of the rectangular piece, and (x2, y2) is the coordinate of the upper-right corner in the original map.

Cases are separated by one blank line.

 

Output

If you can make a complete map with these pieces, output the least number of pieces you need to achieve this. If it is impossible to make one complete map, just output -1.

Sample Input

3
5 5 1
0 0 5 5

5 5 2
0 0 3 5
2 0 5 5

30 30 5
0 0 30 10
0 10 30 20
0 20 30 30
0 0 15 30
15 0 30 30

Sample Output

1
-1
2

Hint

For sample 1, the only piece is a complete map.

For sample 2, the two pieces may overlap with each other, so you can not make a complete treasure map.

For sample 3, you can make a map by either use the first 3 pieces or the last 2 pieces, and the latter approach one needs less pieces.

Author: HANG, Hang
Source: The 6th Zhejiang Provincial Collegiate Programming Contest

 

 

 

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3372

 

 

就是简单的精确覆盖问题。

把每个格子当成一个列，要覆盖所有格子。

 

写一下Dancing Links模板就可以了

 

 

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2014/5/26 21:50:46
  4 File Name     :E:\2014ACM\专题学习\DLX\ZOJ3209.cpp
  5 ************************************************ */
  6 
  7 #include <stdio.h>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <map>
 15 #include <string>
 16 #include <math.h>
 17 #include <stdlib.h>
 18 #include <time.h>
 19 using namespace std;
 20 const int maxnode = 500010;
 21 const int MaxM = 1010;
 22 const int MaxN = 510;
 23 struct DLX
 24 {
 25     int n,m,size;
 26     int U[maxnode],D[maxnode],R[maxnode],L[maxnode],Row[maxnode],Col[maxnode];
 27     int H[MaxN],S[MaxM];
 28     int ansd;
 29     void init(int _n,int _m)
 30     {
 31         n = _n;
 32         m = _m;
 33         for(int i = 0;i <= m;i++)
 34         {
 35             S[i] = 0;
 36             U[i] = D[i] = i;
 37             L[i] = i-1;
 38             R[i] = i+1;
 39         }
 40         R[m] = 0; L[0] = m;
 41         size = m;
 42         for(int i = 1;i <= n;i++)
 43             H[i] = -1;
 44     }
 45     void Link(int r,int c)
 46     {
 47         ++S[Col[++size]=c];
 48         Row[size] = r;
 49         D[size] = D[c];
 50         U[D[c]] = size;
 51         U[size] = c;
 52         D[c] = size;
 53         if(H[r] < 0)H[r] = L[size] = R[size] = size;
 54         else
 55         {
 56             R[size] = R[H[r]];
 57             L[R[H[r]]] = size;
 58             L[size] = H[r];
 59             R[H[r]] = size;
 60         }
 61     }
 62     void remove(int c)
 63     {
 64         L[R[c]] = L[c]; R[L[c]] = R[c];
 65         for(int i = D[c];i != c;i = D[i])
 66             for(int j = R[i];j != i;j = R[j])
 67             {
 68                 U[D[j]] = U[j];
 69                 D[U[j]] = D[j];
 70                 --S[Col[j]];
 71             }
 72     }
 73     void resume(int c)
 74     {
 75         for(int i = U[c];i != c;i = U[i])
 76             for(int j = L[i];j != i;j = L[j])
 77                 ++S[Col[U[D[j]]=D[U[j]]=j]];
 78         L[R[c]] = R[L[c]] = c;
 79     }
 80     void Dance(int d)
 81     {
 82         //剪枝下
 83         if(ansd != -1 && ansd <= d)return;
 84         if(R[0] == 0)
 85         {
 86             if(ansd == -1)ansd = d;
 87             else if(d < ansd)ansd = d;
 88             return;
 89         }
 90         int c = R[0];
 91         for(int i = R[0];i != 0;i = R[i])
 92             if(S[i] < S[c])
 93                 c = i;
 94         remove(c);
 95         for(int i = D[c];i != c;i = D[i])
 96         {
 97             for(int j = R[i];j != i;j = R[j])remove(Col[j]);
 98             Dance(d+1);
 99             for(int j = L[i];j != i;j = L[j])resume(Col[j]);
100         }
101         resume(c);
102     }
103 };
104 DLX g;
105 
106 int main()
107 {
108     //freopen("in.txt","r",stdin);
109     //freopen("out.txt","w",stdout);
110     int T;
111     int n,m,p;
112     scanf("%d",&T);
113     while(T--)
114     {
115         scanf("%d%d%d",&n,&m,&p);
116         g.init(p,n*m);
117         int x1,y1,x2,y2;
118         for(int k = 1;k <= p;k++)
119         {
120             scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
121             for(int i = x1+1;i <= x2;i++)
122                 for(int j = y1+1;j <= y2;j++)
123                     g.Link(k,j + (i-1)*m);
124         }
125         g.ansd = -1;
126         g.Dance(0);
127         printf("%d\n",g.ansd);
128     }
129     return 0;
130 }

 

 

 

 

 


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