题意:4*4的棋盘摆满棋子,有黑有白,翻转一个棋子的同时也将翻转其上下左右的棋子(翻转后黑变白,白变黑),问使棋盘上所有棋子颜色相同,最少翻转的棋子数。
分析:
1、每个棋子至多翻转1次。翻转偶数次与不翻转结果相同,翻转奇数次与翻转1次结果相同。
2、每个棋子翻转或不翻转,共有216种情况。
3、IDA*搜索。
#pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define Min(a, b) ((a < b) ? a : b) #define Max(a, b) ((a < b) ? b : a) const double eps = 1e-8; inline int dcmp(double a, double b){ if(fabs(a - b) < eps) return 0; return a > b ? 1 : -1; } typedef long long LL; typedef unsigned long long ULL; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const LL LL_INF = 0x3f3f3f3f3f3f3f3f; const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {0, -1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const int MAXN = 10000 + 10; const int MAXT = 10000 + 10; using namespace std; char pic[5][5]; char tmp[5][5]; int maxn; bool ok; bool judge_equal(){ for(int i = 0; i < 4; ++i){ for(int j = 0; j < 4; ++j){ if(tmp[i][j] != tmp[0][0]) return false; } } return true; } bool judge(int x, int y){ return x >= 0 && x < 4 && y >= 0 && y < 4; } void flip(int x, int y){ for(int i = 0; i < 5; ++i){ int tmpx = x + dr[i]; int tmpy = y + dc[i]; if(judge(tmpx, tmpy)){ if(tmp[tmpx][tmpy] == 'b') tmp[tmpx][tmpy] = 'w'; else tmp[tmpx][tmpy] = 'b'; } } } void dfs(int x, int y, int cur){ if(cur == maxn){ ok = judge_equal(); return; } if(ok || x == 4) return; flip(x, y); if(y < 3) dfs(x, y + 1, cur + 1); else dfs(x + 1, 0, cur + 1); flip(x, y); if(y < 3) dfs(x, y + 1, cur); else dfs(x + 1, 0, cur); return; } int main(){ for(int i = 0; i < 4; ++i){ scanf("%s", pic[i]); } ok = false; for(maxn = 0; maxn <= 16; ++maxn){ memcpy(tmp, pic, sizeof pic); dfs(0, 0, 0); if(ok) break; } if(!ok) printf("Impossible\n"); else printf("%d\n", maxn); return 0; }