AOJ 0525 - Osenbei

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=50785

翻译：http://blog.csdn.net/acm_10000h/article/details/40988419

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 #include <map>
 #include <queue>

 #define CL(arr, val)    memset(arr, val, sizeof(arr))

 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)

 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("b.txt", "w", stdout);
 #define maxn 1000000000
 #define N 1010
 using namespace std;
 int a[][];
 bool vis[];
 int n,m,ans;

 void dfs(int k)
 {
     if(k==n)
     {
         int top=;
         for(int j=;j<m;j++)
         {
             int cnt=;
             for(int i=;i<n;i++)
             {
                 if(!vis[i]&&a[i][j]||vis[i]&&!a[i][j])  //注意这里，前面和后面只会有一种满足条件，，所以其实统计的只有一种情况。
                     cnt++;
             }
             //printf("%d\n",cnt);
             top+=max(cnt,n-cnt);
         }
         ans=max(top,ans);
         //printf("%d %d\n",top,ans);
         return;
     }
     vis[k]=!vis[k];
     dfs(k+);
     vis[k]=!vis[k];
     dfs(k+);
 }
 int main()
 {
    //Read();
    //Write()
    while(~scanf("%d%d",&n,&m)&&(n+m))
    {
        for(int i=;i<n;i++)
           for(int j=;j<m;j++) scanf("%d",&a[i][j]);
        ans=;
        CL(vis,false);
        dfs();
        printf("%d\n",ans);
    }
    return ;
 }