枚举 POJ 2965 The Pilots Brothers' refrigerator

题目地址：http://poj.org/problem?id=2965

/*
    题意：4*4的矩形，改变任意点，把所有'+'变成'-',，每一次同行同列的都会反转，求最小步数，并打印方案

    DFS：把'+'记为1， '-'记为0
    1. 从(1, 1)走到(4, 4)，每一点DFS两次(改点反转或不反转)；used记录反转的位置
        详细解释：http://poj.org/showmessage?message_id=346723
    2. 比较巧妙的解法：抓住'+'位置反转，'-'位置相对不反转的特点，从状态上考虑
        详细解释：http://poj.org/showmessage?message_id=156561
    3. 枚举步骤数(1~16)，暴力解法，耗时大
        详细解释：http://poj.org/showmessage?message_id=343281
    4. 网上还有其他解法：高斯消元，BFS，+位运算等等

    注意：反转时十字形中心位置多反转了两次，要再反转一次

    我还是DFS写不出来。。。
*/
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <map>
#include <queue>
#include <vector>
using namespace std;

const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int a[5][5];
int used[5][5];

bool ok(void)
{
    for (int i=1; i<=4; ++i)
    {
        for (int j=1; j<=4; ++j)
            if (a[i][j] == 1)    return false;
    }

    return true;
}

void change(int x, int y)
{
    for (int i=1; i<=4; ++i)
    {
        a[x][i] = a[x][i] ? 0 : 1;
        a[i][y] = a[i][y] ? 0 : 1;
    }
    a[x][y] = a[x][y] ? 0 : 1;
    used[x][y] = used[x][y] ? 0 : 1;
}

bool DFS(int x, int y)
{
    if (x == 4 && y == 4)
    {
        if (ok ())    return true;

        change (x, y);
        if (ok ())    return true;

        change (x, y);
        return false;
    }
    int nx, ny;
    if (y == 4)        nx = x + 1, ny = 1;
    else    nx = x, ny = y + 1;

    if (DFS (nx, ny))    return true;

    change (x, y);
    if (DFS (nx, ny))    return true;

    change (x, y);
    return false;
}

void work(void)
{
    if (DFS (1, 1))
    {
        int ans = 0;
        for (int i=1; i<=4; ++i)
        {
            for (int j=1; j<=4; ++j)
            {
                if (used[i][j])        ans++;
            }
        }
        printf ("%d
", ans);
        for (int i=1; i<=4; ++i)
            for (int j=1; j<=4; ++j)
                if (used[i][j])    printf ("%d %d
", i, j);
    }
    else    printf ("WA
");
}

int main(void)        //POJ 2965 The Pilots Brothers' refrigerator
{
    //freopen ("B.in", "r", stdin);

    char ch;
    for (int i=1; i<=4; ++i)
    {
        for (int j=1; j<=4; ++j)
        {
            scanf ("%c", &ch);
            a[i][j] = (ch == '-') ? 0 : 1;
            //printf ("%d ", a[i][j]);
        }
        getchar ();    //puts ("");
    }

    work ();

    return 0;
}

/*
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <map>
#include <queue>
#include <vector>
using namespace std;

const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
bool con[5][5];
int a[5][5];
struct NODE
{
    int x, y;
}node[MAXN];

void work(void)
{
    for (int i=1; i<=4; ++i)
    {
        for (int j=1; j<=4; ++j)
        {
            if (a[i][j] == 1)
            {
                con[i][j] = !con[i][j];
                for (int k=1; k<=4; ++k)
                {
                    con[i][k] = !con[i][k];
                    con[k][j] = !con[k][j];
                }
            }
        }
    }
    int ans = 0;
    for (int i=1; i<=4; ++i)
    {
        for (int j=1; j<=4; ++j)
        {
            if (con[i][j] == true)
            {
                ans++;    node[ans].x = i;    node[ans].y = j;
            }
        }
    }
    printf ("%d
", ans);
    for (int i=1; i<=ans; ++i)
        printf ("%d %d
", node[i].x, node[i].y);
}

int main(void)        //POJ 2965 The Pilots Brothers' refrigerator
{
    //freopen ("B.in", "r", stdin);

    char ch;
    for (int i=1; i<=4; ++i)
    {
        for (int j=1; j<=4; ++j)
        {
            scanf ("%c", &ch);
            a[i][j] = (ch == '-') ? 0 : 1;
        }
        getchar ();
    }

    work ();

    return 0;
}
*/

/*
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <map>
#include <queue>
#include <vector>
using namespace std;

const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int a[5][5];
struct NODE
{
    int x, y;
}node[MAXN];
int k;
bool flag;

bool ok(void)
{
    for (int i=1; i<=4; ++i)
    {
        for (int j=1; j<=4; ++j)
            if (a[i][j] == 1)    return false;
    }

    return true;
}

void change(int x, int y)
{
    for (int i=1; i<=4; ++i)
    {
        a[x][i] = !a[x][i];
        a[i][y] = !a[i][y];
    }
    a[x][y] = !a[x][y];
}

void DFS(int x, int y, int num, int cnt, int kk)
{
    if (num == cnt)
    {
        flag = ok ();
        k = kk;
        return ;
    }
    for (int i=x; i<=4; ++i)
    {
        int j;
        if (i == x)    j = y + 1;
        else    j = 1;
        for (; j<=4; ++j)
        {
            node[kk].x = i;
            node[kk].y = j;
            change (i, j);
            DFS (i, j, num+1, cnt, kk+1);
            if (flag)    return ;
            change (i, j);
        }
    }
}

void work(void)
{
    int cnt;
    for (cnt=1; cnt<=16; ++cnt)
    {
        flag = false;    k = 0;
        DFS (1, 0, 0, cnt, 1);
        if (flag)    break;
    }

    printf ("%d
", cnt);
    for (int i=1; i<=k - 1; ++i)
        printf ("%d %d
", node[i].x, node[i].y);
}

int main(void)        //POJ 2965 The Pilots Brothers' refrigerator
{
    //freopen ("B.in", "r", stdin);

    char ch;
    for (int i=1; i<=4; ++i)
    {
        for (int j=1; j<=4; ++j)
        {
            scanf ("%c", &ch);
            a[i][j] = (ch == '-') ? 0 : 1;
        }
        getchar ();
    }

    work ();

    return 0;
}
*/