POJ

http://poj.org/problem?id=2965

题意：4*4的01方阵，每次翻转一行和一列（交点只翻一次），求把所有面都翻成1的最小方案，多解输出任意一解。

直接暴力模拟：

#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<map>
#include<set>
#include<stack>
#include<string>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;

char cg[4][4];
char g[4][4];

int tx[17], ty[17], ttop;
int ax[17], ay[17], atop;

inline bool less1(int k) {
    int cnt = 0;
    while(k) {
        if(k & 1)
            ++cnt;
        k >>= 1;
    }
    return cnt < atop;
}

inline void flip(int k) {
    ttop = 0;
    int cnt = 0;
    while(k) {
        if(k & 1) {
            ++ttop;
            register int ri = tx[ttop] = cnt / 4;
            register int rj = ty[ttop] = cnt % 4;
            for(register int j = 0; j < 4; ++j) {
                g[ri][j] = !g[ri][j];
            }
            for(register int i = 0; i < 4; ++i) {
                g[i][rj] = !g[i][rj];
            }
            g[ri][rj] = !g[ri][rj];
        }
        k >>= 1;
        ++cnt;
    }
    return;
}

inline bool check() {
    for(int i = 0; i < 4; ++i) {
        for(int j = 0; j < 4; ++j) {
            if(!g[i][j])
                return 0;
        }
    }
    return 1;
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    while(~scanf("%s", cg[0])) {
        for(int j = 0; j < 4; ++j) {
            cg[0][j] = (cg[0][j] == '-');
        }
        for(int i = 1; i < 4; ++i) {
            scanf("%s", cg[i]);
            for(int j = 0; j < 4; ++j) {
                cg[i][j] = (cg[i][j] == '-');
            }
        }
        atop = 17;
        for(int k = 0; k < 1 << 16; ++k) {
            if(less1(k)) {
                memcpy(g, cg, sizeof(g));
                flip(k);
                if(check()) {
                    atop = ttop;
                    memcpy(ax, tx, sizeof(ax));
                    memcpy(ay, ty, sizeof(ay));
                }
            }
        }
        printf("%d
", atop);
        for(int i = 1; i <= atop; ++i) {
            printf("%d %d
", ax[i] + 1, ay[i] + 1);
        }
    }
}

时间有点卡，可能多亏了剪枝才能过。