Rotate Columns (easy version)
题意：
每一列的数字能够进行循环位移，问每一行最大值的和最大是多少
思路：
数据量较小，暴力模拟出每一种情况，然后取最大值即可
一开始想用贪心的思路去做，每次取总体增量最大
然后发现有一种情况会导致错误
第一种：
6 7
6 7
6 7
第二种：
6 6
6 6
6 9
虽然两种增量相同，但是应该是下面这种情况更优，所以错了很久，一开始有想到这种情况，但是脑子转不过来，弄了很久才灵光一闪想到这个样例
1 #include<cstdio> 2 #include<string.h> 3 #include<algorithm> 4 #include<cmath> 5 #include<iostream> 6 #include<vector> 7 #include<queue> 8 #include<set> 9 #include<map> 10 #include<cctype> 11 #define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) 12 #define mem(a,x) memset(a,x,sizeof(a)) 13 #define lson rt<<1,l,mid 14 #define rson rt<<1|1,mid + 1,r 15 #define P pair<int,int> 16 #define ull unsigned long long 17 using namespace std; 18 typedef long long ll; 19 const int maxn = 1e6 + 10; 20 const ll mod = 1000000007; 21 const int inf = 0x3f3f3f3f; 22 const long long INF = 0x3f3f3f3f3f3f3f3f; 23 const double eps = 1e-7; 24 25 inline ll read() 26 { 27 ll X = 0, w = 0; char ch = 0; 28 while (!isdigit(ch)) { w |= ch == '-'; ch = getchar(); } 29 while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar(); 30 return w ? -X : X; 31 } 32 33 int n , m , N; 34 int mp[10][200]; 35 int ans; 36 struct node 37 { 38 int id, maxx; 39 }res[200]; 40 41 int cmp(node& a, node& b) 42 { 43 return a.maxx > b.maxx; 44 } 45 46 47 void dfs(int pos) 48 { 49 if (pos >= N + 1) 50 { 51 int sum = 0; 52 for (int i = 1; i <= n; ++i) 53 { 54 int tmp = 0; 55 for (int j = 1; j <= N; ++j) 56 { 57 tmp = max(tmp, mp[i][res[j].id]); 58 } 59 sum += tmp; 60 } 61 ans = max(ans, sum); 62 } 63 else 64 { 65 int tmp[10]; 66 for (int i = 1; i <= n; ++i) 67 { 68 for (int j = 1; j <= n; ++j) 69 { 70 tmp[j] = mp[j % n + 1][res[pos].id]; 71 } 72 for (int j = 1; j <= n; ++j) 73 { 74 mp[j][res[pos].id] = tmp[j]; 75 } 76 dfs(pos + 1); 77 } 78 } 79 } 80 81 82 void init() 83 { 84 N = 0; 85 mem(res, 0); 86 ans = 0; 87 } 88 89 int main() 90 { 91 int T; 92 T = read(); 93 while (T--) 94 { 95 init(); 96 n = read(), m = read(); 97 for (int i = 1; i <= n; ++i) 98 { 99 for (int j = 1; j <= m; ++j) 100 { 101 mp[i][j] = read(); 102 res[j].id = j; 103 res[j].maxx = max(res[j].maxx, mp[i][j]); 104 } 105 } 106 sort(res + 1, res + 1 + m, cmp); 107 N = min(n, m); 108 dfs(1); 109 cout << ans << endl; 110 } 111 return 0; 112 }