POJ 题目1691 Painting A Board（DFS）

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

Source

Tehran 1999

题目大意：一个区域有n个矩形，给出，每个矩形的左上角坐标和右下角坐标，和它要图的颜色，每个刷子有一种颜色，当只有它上边所有和它挨着的矩形都图了，才能图他，问最少用几个刷子

ac代码

#include<stdio.h>
#include<string.h>
int n;
struct s
{
	int x1,y1,x2,y2,c;
}b[1001];
int map[1010][1010],dig[1010],vis[1010],ans,flag;
void dfs(int dep,int sum,int c)
{
	if(sum>ans)
		return;
	if(dep==n)
	{
	//	flag=1;
		ans=sum;
		return;
	}
	int i,j;
	for(i=0;i<n;i++)
	{
		if(!vis[i]&&dig[i]==0)
		{
			vis[i]=1;
			for(j=0;j<n;j++)
			{
				if(map[i][j])
					dig[j]--;
			}
			if(b[i].c==c)
				dfs(dep+1,sum,c);
			else
				dfs(dep+1,sum+1,b[i].c);
			for(j=0;j<n;j++)
			{
				if(map[i][j])
					dig[j]++;
			}
			vis[i]=0;
		}
	}
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		int i,j;
		for(i=0;i<n;i++)
		{
			scanf("%d%d%d%d%d",&b[i].y1,&b[i].x1,&b[i].y2,&b[i].x2,&b[i].c);
		}
		memset(map,0,sizeof(map));
		memset(dig,0,sizeof(dig));
		memset(vis,0,sizeof(vis));
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
			{
				if(b[i].y2==b[j].y1&&!(b[i].x2<b[j].x1||b[i].x1>b[j].x2))
				{
					dig[j]++;
					map[i][j]=1;
				}
			}
		}
		ans=0xfffffff;
		flag=0;
		dfs(0,0,0);
		printf("%d\n",ans);
	}
}