任意门:http://poj.org/problem?id=3185
The Water Bowls
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 7676 | Accepted: 3036 |
Description
The cows have a line of 20 water bowls from which they drink. The bowls can be either right-side-up (properly oriented to serve refreshing cool water) or upside-down (a position which holds no water). They want all 20 water bowls to be right-side-up and thus use their wide snouts to flip bowls.
Their snouts, though, are so wide that they flip not only one bowl but also the bowls on either side of that bowl (a total of three or -- in the case of either end bowl -- two bowls).
Given the initial state of the bowls (1=undrinkable, 0=drinkable -- it even looks like a bowl), what is the minimum number of bowl flips necessary to turn all the bowls right-side-up?
Their snouts, though, are so wide that they flip not only one bowl but also the bowls on either side of that bowl (a total of three or -- in the case of either end bowl -- two bowls).
Given the initial state of the bowls (1=undrinkable, 0=drinkable -- it even looks like a bowl), what is the minimum number of bowl flips necessary to turn all the bowls right-side-up?
Input
Line 1: A single line with 20 space-separated integers
Output
Line 1: The minimum number of bowl flips necessary to flip all the bowls right-side-up (i.e., to 0). For the inputs given, it will always be possible to find some combination of flips that will manipulate the bowls to 20 0's.
Sample Input
0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0
Sample Output
3
Hint
Explanation of the sample:
Flip bowls 4, 9, and 11 to make them all drinkable:
0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 [initial state]
0 0 0 0 0 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 [after flipping bowl 4]
0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 [after flipping bowl 9]
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [after flipping bowl 11]
Flip bowls 4, 9, and 11 to make them all drinkable:
0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 [initial state]
0 0 0 0 0 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 [after flipping bowl 4]
0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 [after flipping bowl 9]
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [after flipping bowl 11]
Source
题意概括:
N个开关,打开一个开关相邻的开关状态会取反,给一个初始的所有开关状态,要求求出最小的改变开关的次数使得所有开关的状态为关闭;
解题思路:
构造增广矩阵类似于根据开关的关系构造有向图的邻接矩阵;
构造增广矩阵,高斯消元,枚举自由元(二进制枚举状态),寻找最小值;
AC code:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <cstdio> 2 #include <iostream> 3 #include <algorithm> 4 #include <cstring> 5 #include <cmath> 6 #define INF 0x3f3f3f3f 7 #define LL long long 8 using namespace std; 9 const int MAXN = 300; 10 int a[MAXN][MAXN]; //增广矩阵 11 int freeX[MAXN]; //自由元 12 int x[MAXN]; //解集 13 int equ, var; 14 int free_num; 15 int N; 16 17 int Gauss() 18 { 19 int maxRow, col, k; 20 free_num = 0; 21 for(k = 0, col = 0; k < equ && col < var; k++, col++){ 22 maxRow = k; 23 for(int i = k+1; i < equ; i++){ 24 if(abs(a[i][col]) > abs(a[maxRow][col])){ 25 maxRow = i; 26 } 27 } 28 29 if(a[maxRow][col] == 0){ 30 k--; 31 freeX[free_num++] = col; 32 continue; 33 } 34 if(maxRow != k){ 35 for(int j = col; j < var+1; j++){ 36 swap(a[k][j], a[maxRow][j]); 37 } 38 } 39 40 for(int i = k+1; i < equ; i++){ 41 if(a[i][col] != 0){ 42 for(int j = col; j < var+1; j++) 43 a[i][j] ^= a[k][j]; 44 45 } 46 } 47 } 48 49 for(int i = k; i < equ; i++) //无解 50 if(a[i][col] != 0) return -1; 51 52 if(k < var) return var-k; //多解返回自由元个数 53 54 for(int i = var-1; i >= 0; i--){ //唯一解,回代 55 x[i] = a[i][var]; 56 for(int j = i+1; j < var; j++){ 57 x[i] ^= (a[i][j] && x[j]); 58 } 59 } 60 return 0; 61 } 62 63 64 void solve() 65 { 66 int t = Gauss(); 67 if(t == -1){ //无解的情况,其实题目保证有解 68 printf("inf "); 69 return; 70 } 71 else if(t == 0){ //唯一解 72 int ans = 0; 73 for(int i = 0; i < N; i++){ 74 ans += x[i]; 75 } 76 printf("%d ", ans); 77 return; 78 } 79 else{ //多解,枚举自由元 80 int ans = INF; 81 int tot = (1<<t); 82 for(int i = 0; i < tot; i++){ 83 int cnt = 0; 84 for(int j = 0; j < t; j++){ 85 if(i&(1<<j)){ 86 x[freeX[j]] = 1; 87 cnt++; 88 } 89 else x[freeX[j]] = 0; 90 } 91 92 for(int j = var-t-1; j >= 0; j--){ 93 int index; 94 for(index = j; index < var; index++) 95 if(a[j][index]) 96 break; 97 x[index] = a[j][var]; 98 99 for(int s = index+1; s < var; s++) 100 if(a[j][s]) 101 x[index] ^= x[s]; 102 cnt += x[index]; 103 } 104 ans = min(ans, cnt); 105 } 106 printf("%d ", ans); 107 } 108 return; 109 } 110 111 int main() 112 { 113 N = 20; 114 equ = 20; 115 var = 20; 116 memset(a, 0, sizeof(a)); 117 memset(x, 0, sizeof(x)); 118 for(int i = 0; i < N; i++){ 119 a[i][i] = 1; 120 if(i > 0) a[i-1][i] = 1; 121 if(i < N-1) a[i+1][i] = 1; 122 } 123 for(int i = 0; i < N; i++){ 124 scanf("%d", &a[i][N]); 125 } 126 solve(); 127 return 0; 128 }