POJ 1681 Painter's Problem 【高斯消元 二进制枚举】

  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\n");
 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\n", 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\n", 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 }