HDU 4890 One to Four（2014 Multi-University Training Contest 3）

题意：给定一个长方形网格，要把它切成完全相同4个部分（这里完全相同指可以旋转平移后能重叠）。把4个重叠后每个网格对应有四个数字相加，得到一种方案，所有格子中和最小就是该种方案的值，在多种方案中，最后问最大的解能使多少。

思路：首先任意一种划分可以映射到4个相同的长方形， 也就是说一种不是长方形的方案 可以转化为一种长方形的。那么我们就只需要找4个完全相同的长方形就可以。一共有以下画法：

下面画图时基于n>m。边长要满足一定条件要自己注意。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define debug(x) printf(#x"= %d\n",x);
using namespace std;
struct node {
    int a[][];
    int n, m;
    void rota() {//旋转90度
        int b[][];
        for (int i = ; i < n; ++i) {
            for (int j = ; j < m; ++j) {
                b[j][n - i - ] = a[i][j];
            }
        }
        swap(n, m);
        for (int i = ; i < n; ++i)
            for (int j = ; j < m; ++j)
                a[i][j] = b[i][j];
    }
} p[], s;
int a[][];
int n, m;
int ans;
node cur[];
void gao(int now) {
    if (now == ) {
        for (int i = ; i < ; ++i)
            if (cur[i].n != cur[i + ].n || cur[i].m != cur[i + ].m)
                return;

        int minn = ;
        for (int i = ; i < cur[].n; ++i) {
            for (int j = ; j < cur[].m; ++j) {
                int sum = ;
                for (int k = ; k < ; ++k) {
                    sum += cur[k].a[i][j];
                }
                //    printf("%d %d %d\n",i,j,sum);
                if (sum < minn)
                    minn = sum;
            }
        }
        if (minn > ans)
            ans = minn;

        return;
    }
    cur[now] = p[now];
    gao(now + );
    for (int i = ; i < ; ++i) {
        cur[now].rota();
        gao(now + );
    }
}
int main() {
    while (scanf("%d%d", &n, &m) != EOF) {
        for (int i = ; i < n; ++i)
            for (int j = ; j < m; ++j)
                scanf("%d", &a[i][j]);
        //debug(m);
        ans = ;
        if (n %  == ) {
            for (int i = ; i < ; ++i) {
                for (int j = n /  * i; j < n /  * (i + ); ++j) {
                    for (int k = ; k < m; ++k) {
                        p[i].a[j % (n / )][k] = a[j][k];
                    }
                }
                p[i].n = n / ;
                p[i].m = m;
            }
            gao();
        }

        if (m %  == ) {
            for (int i = ; i < ; ++i) {
                for (int j = ; j < n; ++j) {
                    for (int k = i * (m / ); k < (i + ) * m / ; ++k) {
                        p[i].a[j][k % (m / )] = a[j][k];
                    }
                }
                p[i].n = n;
                p[i].m = m / ;
            }
            gao();
        }

        if (n %  ==  && m %  == ) {
            for (int i = ; i < ; ++i) {
                for (int j = (i / ) * n / ; j < (i /  + ) * n / ; ++j) {
                    for (int k = (i % ) * m / ; k < (i %  + ) * m / ;
                            ++k) {
                        p[i].a[j % (n / )][k % (m / )] = a[j][k];
                    }
                }
                p[i].n = n / ;
                p[i].m = m / ;
            }
            gao();
        }

        s.n = n;
        s.m = m;
        for (int i = ; i < n; ++i)
            for (int j = ; j < m; ++j)
                s.a[i][j] = a[i][j];
        if (n < m) {
            s.rota();
            swap(n, m);
        }
        if ( * s.n ==  * s.m && s.m %  ==  && s.n %  == ) {
            for (int i = ; i < ; ++i) {
                for (int j = ; j < m; ++j) {
                    for (int k = i * m / ; k < (i + ) * m / ; ++k) {
                        p[i].a[j][k % (m / )] = s.a[j][k];
                    }
                }
                p[i].n = m;
                p[i].m = m / ;
            }
            for (int i = m; i < n; ++i)
                for (int j = ; j < m; ++j)
                    p[].a[i - m][j] = s.a[i][j];
            p[].n = n - m;
            p[].m = m;
            gao();

            for (int i = ; i < ; ++i) {
                for (int j = n - m; j < n; ++j) {
                    for (int k = i * m / ; k < (i + ) * m / ; ++k) {
                        p[i].a[j - (n - m)][k % (m / )] = s.a[j][k];
                    }
                }
                p[i].n = m;
                p[i].m = m / ;
            }
            for (int i = ; i < n - m; ++i)
                for (int j = ; j < m; ++j)
                    p[].a[i][j] = s.a[i][j];
            p[].n = n - m;
            p[].m = m;
            gao();
        }

        if (s.n ==  * s.m && s.m %  == ) {
            for (int i = ; i < n / ; ++i)
                for (int j = ; j < m / ; ++j)
                    p[].a[i][j] = s.a[i][j];
            p[].n = s.n / ;
            p[].m = s.m / ;
            for (int i = ; i < n / ; ++i)
                for (int j = m / ; j < m; ++j)
                    p[].a[i][j - m / ] = s.a[i][j];
            p[].n = s.n / ;
            p[].m = s.m / ;
            for (int i = n / ; i < n /  * ; ++i)
                for (int j = ; j < m; ++j)
                    p[].a[i - n / ][j] = s.a[i][j];
            p[].n = s.n / ;
            p[].m = s.m;
            for (int i = n /  * ; i < n; ++i)
                for (int j = ; j < m; ++j)
                    p[].a[i - n /  * ][j] = s.a[i][j];
            p[].n = n / ;
            p[].m = m;
            gao();

            for (int i = n / ; i < n /  * ; ++i)
                for (int j = ; j < m / ; ++j)
                    p[].a[i - n / ][j] = s.a[i][j];
            p[].n = s.n / ;
            p[].m = s.m / ;
            for (int i = n / ; i < n /  * ; ++i)
                for (int j = m / ; j < m; ++j)
                    p[].a[i - n / ][j - m / ] = s.a[i][j];
            p[].n = s.n / ;
            p[].m = s.m / ;
            for (int i = ; i < n / ; ++i)
                for (int j = ; j < m; ++j)
                    p[].a[i][j] = s.a[i][j];
            p[].n = s.n / ;
            p[].m = s.m;
            for (int i = n /  * ; i < n; ++i)
                for (int j = ; j < m; ++j)
                    p[].a[i - n /  * ][j] = s.a[i][j];
            p[].n = n / ;
            p[].m = m;
            gao();

            for (int i = n / ; i < n; ++i)
                for (int j = ; j < m / ; ++j)
                    p[].a[i - n / ][j] = s.a[i][j];
            p[].n = s.n / ;
            p[].m = s.m / ;
            for (int i = n / ; i < n; ++i)
                for (int j = m / ; j < m; ++j)
                    p[].a[i - n / ][j - m / ] = s.a[i][j];
            p[].n = s.n / ;
            p[].m = s.m / ;
            for (int i = ; i < n / ; ++i)
                for (int j = ; j < m; ++j)
                    p[].a[i][j] = s.a[i][j];
            p[].n = s.n / ;
            p[].m = s.m;
            for (int i = n / ; i < n / ; ++i)
                for (int j = ; j < m; ++j)
                    p[].a[i - n / ][j] = s.a[i][j];
            p[].n = n / ;
            p[].m = m;
            gao();

        }
        printf("%d\n", ans);

    }
    return ;
}