【poj 1691】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：一个矩阵中，给出很多个矩形块，每个块有一个期待颜色，让你给块染色，规则是：如果要染一个块，那么他上面所有的块都必须已经被染色,并且每次换一个颜色时，都要换一个刷子

(比如我的染色顺序为（编号代表颜色号）１，２，１，那么要用３个刷子，但如果是１，１，２，那么只要２个刷子)

　　求染完所有块所需最小刷子数

解：很明显的拓扑思想，网上有的题意题解太复杂

1.建图：建一条有向边从上方矩形指向相邻的下方矩形

模仿拓扑排序，in[]数组记录每个点入度

2.搜索:每次搜索条件是:入度为0,即上方所有矩形已经染　　　　　　　　　　色，并且没有被染色过

细节不多，个人觉得建图比较难，代码如下：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
struct node{int x1,x2,y1,y2;int id,c;} arr[30];
vector<int> lin[30];
int ans,n,T,in[30],vis[30];
int dfs(int tot,int step,int pre)
{
if (step>=ans)  return 0;//简单的剪枝
if (tot==n)
{
ans=step;return 0;
}
for (int i=1;i<=n;i++)//枚举每个能搜索的点
if (vis[i]==0 && in[i]==0)
{
if (arr[i].c==pre)　//如果上个颜色和这次一样，不用新的刷子
{
vis[i]=1;
for (int j=0;j<lin[i].size();j++)
in[lin[i][j]]--;
dfs(tot+1,step,pre);
vis[i]=0;
for (int j=0;j<lin[i].size();j++)
in[lin[i][j]]++;
continue;
}
else//如果上个颜色和这次不一样，刷子数＋＋
{
vis[i]=1;
for (int j=0;j<lin[i].size();j++)
in[lin[i][j]]--;
dfs(tot+1,step+1,arr[i].c);
vis[i]=0;
for (int j=0;j<lin[i].size();j++)
in[lin[i][j]]++;
continue;
}
}
}

int main()
{
cin>>T;
while(T--)
{
scanf("%d",&n);
ans=99999999;
for (int i=1;i<=n;i++)
{
scanf("%d%d%d%d%d",&arr[i].y1,&arr[i].x1,&arr[i].y2,&arr[i].x2,&arr[i].c);
arr[i].id=i;
lin[i].clear();
vis[i]=0;
in[i]=0;
}
//建图
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
if (arr[j].x1<arr[i].x2 && arr[j].x2>arr[i].x1 &&
arr[j].y1==arr[i].y2)
{
lin[i].push_back(j);
in[j]++;
}
//枚举每个能开始的点开始搜索
for (int i=1;i<=n;i++)
if (in[i]==0 && vis[i]==0)
{
vis[i]=1;
for (int j=0;j<lin[i].size();j++)
in[lin[i][j]]--;
dfs(1,1,arr[i].c);
vis[i]=0;
for (int j=0;j<lin[i].size();j++)
in[lin[i][j]]++;
}
printf("%d\n",ans);
}
}