[POJ1753]Flip Game（开关问题，枚举）

题目链接：http://poj.org/problem?id=1753

和上一个题一样，将初始状态存成01矩阵，就可以用位运算优化了。黑色白色各来一遍

 /*
 ━━━━━┒ギリギリ♂ eye！
 ┓┏┓┏┓┃キリキリ♂ mind！
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┃┃┃┃┃┃
 ┻┻┻┻┻┻
 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%lld", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f, sizeof(a))
 #define lp p << 1
 #define rp p << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 const int maxn = ;
 const int dx[] = {, , , , -};
 const int dy[] = {, , -, , };
 int G[maxn][maxn];
 int tmp[maxn][maxn];
 int ret[maxn][maxn];
 int n, m, ans;
 char oho[maxn][maxn];

 bool ok(int x, int y) {
     return x >=  && x < m && y >=  && y < n;
 }

 int get(int x, int y) {
     int c = G[x][y];
     Rep(i, ) {
         int xx = x + dx[i];
         int yy = y + dy[i];
         if(ok(xx, yy)) c += tmp[xx][yy];
     }
     return c % ;
 }

 int calc() {
     For(i, , m) Rep(j, n) if(get(i-, j) != ) tmp[i][j] = ;
     Rep(i, n) if(get(m-, i) != ) return -;
     int p = ;
     Rep(i, m) Rep(j, n) p += tmp[i][j];
     return p;
 }

 int main() {
     // FRead();
     n = m = ;
     Cls(oho); Cls(ret); Cls(tmp); Cls(G); ans = -;
     Rep(i, n) Rs(oho[i]);
     Rep(i, m) {
         Rep(j, n) {
             if(oho[i][j] == 'b') G[i][j] = ;
             else G[i][j] = ;
         }
     }
     int nn =  << n;
     Rep(i, nn) {
         Cls(tmp);
         Rep(j, n) tmp[][n-j-] = i >> j & ;
         int num = calc();
         if(num >=  && (ans <  || ans > num)) ans = num;
     }
     Cls(ret); Cls(tmp); Cls(G);
     Rep(i, m) {
         Rep(j, n) {
             if(oho[i][j] == 'b') G[i][j] = ;
             else G[i][j] = ;
         }
     }
     Rep(i, nn) {
         Cls(tmp);
         Rep(j, n) tmp[][n-j-] = i >> j & ;
         int num = calc();
         if(num >=  && (ans <  || ans > num)) ans = num;
     }
     if(ans < ) puts("Impossible");
     else printf("%d\n", ans);
     RT ;
 }