POJ 1681 Painter's Problem 高斯消元 二进制枚举

Posted 2021-01-20 ymzjj

任意门：http://poj.org/problem?id=1681

Painter‘s Problem

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7667		Accepted: 3624

Description

There is a square wall which is made of n*n small square bricks. Some bricks are white while some bricks are yellow. Bob is a painter and he wants to paint all the bricks yellow. But there is something wrong with Bob‘s brush. Once he uses this brush to paint brick (i, j), the bricks at (i, j), (i-1, j), (i+1, j), (i, j-1) and (i, j+1) all change their color. Your task is to find the minimum number of bricks Bob should paint in order to make all the bricks yellow. 

Input

The first line contains a single integer t (1 <= t <= 20) that indicates the number of test cases. Then follow the t cases. Each test case begins with a line contains an integer n (1 <= n <= 15), representing the size of wall. The next n lines represent the original wall. Each line contains n characters. The j-th character of the i-th line figures out the color of brick at position (i, j). We use a ‘w‘ to express a white brick while a ‘y‘ to express a yellow brick.

Output

For each case, output a line contains the minimum number of bricks Bob should paint. If Bob can‘t paint all the bricks yellow, print ‘inf‘.

Sample Input

2
3
yyy
yyy
yyy
5
wwwww
wwwww
wwwww
wwwww
wwwww

Sample Output

0
15

Source

POJ Monthly--2004.06.27 张嘉龄

 

题意概括：

一个二维矩阵， 输入每个格子的初始颜色，可以进行的操作是 粉刷一个格子则相邻的上下左右四个各自颜色都会取反（只有两种颜色）；

问最后把全部各自涂成黄色的最小操作数；

解题思路：

根据题意，每个各自都是一个变元，根据各自之间的相邻关系构造增广矩阵；

高斯消元求出自由元个数 sum；

二进制枚举方案，找出最小的操作数;

 

AC code：

  1 #include <cstdio>
  2 #include <iostream>
  3 #include <algorithm>
  4 #include <cstring>
  5 #define INF 0x3f3f3f3f
  6 #define LL long long
  7 using namespace std;
  8 const int MAXN = 300;
  9 
 10 int equ, var;
 11 int a[MAXN][MAXN];
 12 char str[MAXN][MAXN];
 13 int x[MAXN];
 14 int free_x[MAXN];
 15 int free_num;
 16 int N;
 17 
 18 int Gauss()
 19 {
 20     int maxRow, col, k;
 21     free_num = 0;
 22     for(k = 0, col = 0; k < equ && col < var; k++, col++){
 23         maxRow = k;
 24         for(int i = k+1; i <equ; i++){
 25             if(abs(a[i][col]) > abs(a[maxRow][col]))
 26                 maxRow = i;
 27         }
 28         if(a[maxRow][col] == 0){                //最大的都为0说明该列下面全是 0
 29             k--;
 30             free_x[free_num++] = col;           //说明col是自由元
 31             continue;
 32         }
 33         if(maxRow != k){                        //交换行
 34             for(int j = col; j < var+1; j++)
 35                 swap(a[k][j], a[maxRow][j]);
 36         }
 37         for(int i = k+1; i < equ; i++){             //消元
 38             if(a[i][col] != 0){
 39                 for(int j = col; j < var+1; j++){
 40                     a[i][j] ^= a[k][j];
 41                 }
 42             }
 43         }
 44     }
 45     for(int i = k; i< equ; i++){
 46         if(a[i][col] != 0) return -1;   //无解
 47     }
 48     if(k < var) return var-k;           //返回自由元个数
 49 
 50     for(int i = var-1; i >= 0; i--){    //唯一解，回代
 51         x[i] = a[i][var];
 52         for(int j = i+1; j < var; j++){
 53             x[i] ^= (a[i][j] && x[j]);
 54         }
 55     }
 56     return 0;
 57 }
 58 
 59 void init()
 60 {
 61     memset(a, 0, sizeof(a));
 62     memset(x, 0, sizeof(x));
 63     equ = N*N;
 64     var = N*N;
 65     for(int i = 0; i < N; i++){                 //构造增广矩阵
 66         for(int j = -0; j < N; j++){
 67             int t = i*N+j;
 68             a[t][t] = 1;
 69             if(i > 0) a[(i-1)*N+j][t] = 1;
 70             if(i < N-1) a[(i+1)*N+j][t] = 1;
 71             if(j > 0) a[i*N+j-1][t] = 1;
 72             if(j < N-1) a[i*N+j+1][t] = 1;
 73         }
 74     }
 75 }
 76 
 77 void solve()
 78 {
 79     int t = Gauss();
 80     if(t == -1){            //无解
 81         printf("inf
");
 82         return;
 83     }
 84     else if(t == 0){        //唯一解
 85         int ans = 0;
 86         for(int i = 0; i < N*N; i++){
 87             ans += x[i];
 88         }
 89         printf("%d
", ans);
 90         return;
 91     }
 92     else{                   //多解，需要枚举自由元
 93         int ans = INF;
 94         int tot = 1<<t;
 95         int cnt = 0;
 96         for(int i = 0; i < tot; i++){
 97             cnt = 0;
 98             for(int j = 0; j < t; j++){
 99                 if(i&(1<<j)){
100                     x[free_x[j]] = 1;
101                     cnt++;
102                 }
103                 else x[free_x[j]] = 0;
104             }
105 
106             for(int j = var-t-1; j >= 0; j--){
107                 int index;
108                 for(index = j; index < var; index++){
109                     if(a[j][index]) break;
110                 }
111                 x[index] = a[j][var];
112 
113                 for(int s = index+1; s < var; s++)
114                 if(a[j][s]) x[index] ^= x[s];
115 
116                 cnt+=x[index];
117             }
118             ans = min(ans, cnt);
119         }
120         printf("%d
", ans);
121     }
122 }
123 
124 int main()
125 {
126     int T_case;
127     int tpx, tpy;
128     scanf("%d", &T_case);
129     while(T_case--){
130         scanf("%d", &N);
131         init();
132         for(int i = 0; i < N; i++){
133             scanf("%s", str[i]);
134             for(int j = 0; j < N; j++){
135                 tpx = i*N+j, tpy = N*N;
136                 if(str[i][j] == ‘y‘) a[tpx][tpy] = 0;
137                 else a[tpx][tpy] = 1;
138             }
139         }
140         solve();
141     }
142     return 0;
143 }

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