poj 1691 Painting A Board

Painting A Board

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3018		Accepted: 1441

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
题目大概意思就是有n个矩形需要涂上不同的颜色，问换尽量少的刷子时，将所有的矩形涂上颜色， 一个矩形可以上颜色当且仅当在它上面的矩形都涂上了颜色

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std ;
#define N 16
int dp[1<<N][N] ; // dp[s][i]  表示在状态s下第i个矩形上颜色的换最少的刷子
int state[N]; // state[i] 表示第i个矩形要涂上颜色时必须满足它上面的矩形也要涂上矩形

struct Point{
    int x, y ;
} ;

struct Rec{
    Point p1, p2 ;
    int c ;
} r[N];

void prework(int n){
    for(int i=0; i<n; i++){
        for(int j=0; j<n; j++){
            if(r[j].p2.x <= r[i].p1.x && (!(r[j].p1.y > r[i].p2.y || r[j].p2.y < r[i].p1.y)))
                state[i] |= (1<<j) ;
        }
    }
}
int main(){
    //freopen("input.txt", "r", stdin) ;
    int T ;
    scanf("%d", &T) ;
    while(T--){
        memset(state, 0, sizeof(state)) ;
        int n ;
        scanf("%d", &n) ;
        for(int i=0; i<n; i++)  scanf("%d %d %d %d %d", &r[i].p1.x, &r[i].p1.y, &r[i].p2.x, &r[i].p2.y, &r[i].c) ;
        prework(n) ;
        for(int i=0; i<n; i++)
            for(int j=1; j<(1<<n); j++)
                dp[j][i] = 20 ;
        for(int s=1; s<(1<<n); s++){
            for(int i=0; i<n; i++){
                if(s & (1<<i)){
                    if((s & state[i]) == state[i]){
                        if(s == (1<<i))   dp[s][i] = 1 ;
                        else{
                            for(int j=0; j<n; j++){
                                if(j == i || !(s & (1<<j))) continue ;
                                dp[s][i] = min(dp[s][i], dp[s^(1<<i)][j] + (r[j].c != r[i].c)) ;
                            }
                        }
                    }
                }
            }
        }
        int ans = 20 ;
        for(int i=0; i<n; i++)  ans = min(ans, dp[(1<<n)-1][i]) ;
        printf("%d\n", ans) ;
    }
    return 0 ;
}