POJ-1691 Painting A Board
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 3616 | Accepted: 1798 |
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.
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:
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
#include <cstdio>
#include <cstring>
#include <iostream>
#define MAXN 2147483647
using namespace std;
int f[16][16],d[16],x1[16],x2[16],y1[16],y2[16],col[16],ans,n,m;
bool done[16];
bool jud(int i,int j)
{
if(x1[i] <= x1[j] && x1[j] <= x2[i]) return true;
if(x1[i] <= x2[j] && x2[j] <= x2[j]) return true;
return false;
}
void dfs(int k,int tot,int d[16],bool done[16])
{
int Tot,D[16];
bool DONE[16];
if(k >= ans) return;
if(tot == n)
{
ans=k;
return;
}
for(int i = 1;i <= 20;i++)
{
for(int j = 1;j <= n;j++)
if(!done[j] && d[j] == 0 && col[j] == i)
{
bool flag;
Tot=tot;
for(int I = 1;I <= n;I++) D[I]=d[I];
for(int I = 1;I <= n;I++) DONE[I]=done[I];
do
{
flag=false;
for(int now = 1;now <= n;now++)
if(!DONE[now] && D[now] == 0 && col[now] == i)
{
DONE[now]=true;
Tot++;
for(int l = 1;l <= n;l++)
if(f[now][l]) D[l]--;
flag=true;
}
} while(!flag);
dfs(k+1,Tot,D,DONE);
break;
}
}
}
int main()
{
cin.sync_with_stdio(false);
cin>>m;
for(int t = 1;t <= m;t++)
{
memset(f,0,sizeof(f));
memset(d,0,sizeof(d));
cin>>n;
for(int i = 1;i <= n;i++) cin>>y1[i]>>x1[i]>>y2[i]>>x2[i]>>col[i];
for(int i = 1;i <= n;i++)
for(int j = 1;j <= n;j++)
if(i - j)
if( y1[j] == y2[i] && jud(i,j) )
{
f[i][j]=1;
d[j]++;
}
ans=MAXN;
dfs(1,0,d,done);
cout<<ans-1<<endl;
}
}