【POJ 1691】 Painting A Board（dfs）

[align=center]【POJ 1691】 Painting A Board（dfs）[/align]

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3582		Accepted: 1781

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

Source
Tehran 1999

题目大意:有一块画布，被分割成了n块矩形，每块矩形有自己的颜色，可能有某些部分颜色相同。

现在要对画布涂色，对应位置要涂成对应的颜色。

对要进行涂色的矩形有如下要求:在它上方的矩形必须已经被涂完。

每当选择一种颜色时，可以对所有要涂这种颜色且可以涂色的区域涂色。当切换另一种眼色时，上一种颜色会被清洗掉。

问最少需要多少色料(每次涂同种颜色时用掉一份色料

由于颜色最多只有20种。

可以把所有矩形区域存下，然后dfs搜索所有涂色顺序。

并且要记得对矩形排个序，以免先访问下面的矩形，可能此时无法对其涂色，但访问到其上面的矩形时，如果能涂色，该矩形也就能涂色了。

所以要提前从上往下按坐标排个序。

对于涂色，没有必要整个矩形区域涂色，只需要涂它的下界。因为只有这个才是有用的标记(下方的矩形需要依靠上方矩形的下界来判断是否可以被涂色

代码如下：

#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define Pr pair<int,int>
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)

using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const int mod = 1e9+7;
const double eps = 1e-8;

struct Range
{
int x1,y1,x2,y2,c;
bool f;
bool operator < (const struct Range r)const
{
return x1 == r.x1? y1 < r.y1: x1 < r.x1;
}
};

Range rg[33];
bool mp[111][111];
int n,mn;

void col(int pos)
{
for(int i = rg[pos].y1; i <= rg[pos].y2; ++i)
mp[rg[pos].x2][i] = 1;
}

bool can(int pos)
{
if(rg[pos].x1 == 0) return true;

for(int i = rg[pos].y1; i <= rg[pos].y2; ++i)
if(!mp[rg[pos].x1][i]) return false;

return true;
}

bool cal(int c,int *tmp,int &tp)
{
tp = 0;

for(int i = 0; i < n; ++i)
{
if(rg[i].f || rg[i].c != c) continue;
if(can(i))
{
rg[i].f = 1;
tmp[tp++] = i;
col(i);
}
}

return tp != 0;
}

void reset(int *tmp,int tp)
{
for(int i = 0; i < tp; ++i)
{
rg[tmp[i]].f = 0;
for(int j = rg[tmp[i]].y1; j <= rg[tmp[i]].y2; ++j)
mp[rg[tmp[i]].x2][j] = 0;
}
}

void dfs(int ok,int us)
{
if(ok == n)
{
mn = min(mn,us);
return;
}

int tmp[33];
int tp;
for(int i = 1; i <= 20; ++i)
{
if(!cal(i,tmp,tp)) continue;
//printf("%d->%d %d %d\n",ok,ok+tp,us,i);
dfs(ok+tp,us+1);
reset(tmp,tp);
}

}

int main()
{
//fread();
//fwrite();

int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i = 0; i < n; ++i)
{
scanf("%d%d%d%d%d",&rg[i].x1,&rg[i].y1,&rg[i].x2,&rg[i].y2,&rg[i].c);
rg[i].f = 0;
}
sort(rg,rg+n);
mn = INF;
memset(mp,0,sizeof(mp));
dfs(0,0);
printf("%d\n",mn);
}

return 0;
}