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:

1. Color-code is an integer in the range of 1 .. 20.
   
2. Upper left corner of the board coordinates is always (0,0).
   
3. Coordinates are in the range of 0 .. 99.
   
4. 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);
           
            vis[i] = false;
            for(int j = 0; j < n; ++j)
            {
                if(map[i][j]) //补回小矩阵i的后继
                    deg[j]++;
            }
        }
    }
}

int main()
{
//    freopen("input.txt","r",stdin);
    int caseNum;
    scanf("%d",&caseNum);
    while(caseNum--)
    {
        memset(map,false,sizeof(map));
        memset(deg,0,sizeof(deg));
        memset(vis,false,sizeof(vis));
        scanf("%d",&n);
        for(int i = 0; i < n; ++i)
            scanf("%d %d %d %d %d",&rec[i].y1,&rec[i].x1,&rec[i].y2,&rec[i].x2,&rec[i].color);
        result = INF;
        read_graph();
        dfs(0,0,0);
        printf("%d\n",result);
    }
    return 0;
}