ACM: 搜索题 poj 1691 一觉醒来居…

                                                        
Painting A Board

Description
The CE digital
company has built an Automatic Painting Machine (APM) to paint a
flat board fully covered by adjacent non-overlapping rectangles of
different sizes each with a predefined color.



To color the board, the APM has access to a set of brushes. Each
brush has a distinct color C. The APM picks one brush with color C
and paints all possible rectangles having predefined color C with
the following restrictions:

To avoid leaking the paints and mixing colors, a rectangle can only
be painted if all rectangles immediately above it have already been
painted. For example rectangle labeled F in Figure 1 is painted
only after rectangles C and D are painted. Note that each rectangle
must be painted at once, i.e. partial painting of one rectangle is
not allowed.

You are to write a program for APM to paint a given board so that
the number of brush pick-ups is minimum. Notice that if one brush
is picked up more than once, all pick-ups are counted.

Input

The first line of
the input file contains an integer M which is the number of test
cases to solve (1 <= M <= 10). For
each test case, the first line contains an integer N, the number of
rectangles, followed by N lines describing the rectangles. Each
rectangle R is specified by 5 integers in one line: the y and x
coordinates of the upper left corner of R, the y and x coordinates
of the lower right corner of R, followed by the color-code of
R.

Note that:
Color-code is an integer in the range of 1 .. 20.

Upper left corner of the board coordinates is always
(0,0).

Coordinates are in the range of 0 .. 99.

N is in the range of 1..15.

Output

One line for each
test case showing the minimum number of brush pick-ups.
Sample Input

1

7

0 0 2 2 1

0 2 1 6 2

2 0 4 2 1

1 2 4 4 2

1 4 3 6 1

4 0 6 4 1

3 4 6 6 2

Sample Output

3

题意: 现在一个自动画板要用最少的画刷次数完成任务,
有一个要求就是: 

        
每个小矩阵要染色的时候必须要先将 与其相邻和上面的小矩阵染色.

        
每个小矩阵给出左上角坐标和右下角坐  标和颜色的编号.

解题思路:

             
1. 染色问题. 先将图建立起来. 将当前小矩阵编号为i, 与其相邻的或则在上面的矩阵链接起来.(做标记)

             
2. 深搜解决. dfs(int reLen,int sum,int color)

                   
reLen: 小矩阵已经染色的数目, sum: 当前使用画刷的次数. color: 当前画刷的颜色.

代码:

#include
<cstdio>

#include <iostream>

#include <cstring>

using namespace std;

#define MAX 20

const int INF = (1<<30);

struct node

{

    int x1,
y1;

    int x2,
y2;

    int
color;

}rec[MAX];

int n;

bool vis[MAX], map[MAX][MAX];

int deg[MAX];

int result;

void read_graph()

{

    for(int i =
0; i < n; ++i)

    {

   
    for(int j =
0; j < n; ++j)

   
    {

   
   
    if(rec[i].y2
== rec[j].y1 && (rec[j].x1
<= rec[i].x2 &&
rec[i].x1 <= rec[j].x2))

   
   
    {

   
   
   
    deg[j]++;
//第j个小矩阵有多少个前躯.

   
   
   
    map[i][j] =
true;

   
   
    }

   
    }

    }

}

void dfs(int reLen,int sum,int color)

{

    if(sum
> result) return ;

    if(reLen ==
n)

    {

   
    result =
sum;

   
    return
;

    }

   

    for(int i =
0; i < n; ++i)

    {

   
    if(!vis[i]
&& deg[i] == 0)
//每次要从当前没有前躯的小矩阵开始染色.

   
    {

   
   
    vis[i] =
true;

   
   
    for(int j =
0; j < n; ++j)

   
   
    {

   
   
   
   
if(map[i][j]) //把小矩阵i的后继全部删除.

   
   
   
   
   
deg[j]--;

   
   
    }

   
   
   

   
   
   
if(rec[i].color == color)

   
   
   
   
dfs(reLen+1,sum,color);

   
   
    else

   
   
   
   
dfs(reLen+1,sum+1,rec[i].color);