如果暴力的话，时间复杂度是\(rcC(n, n/2)^2\)，主要考察搜索枚举行和列，并没有用到DP的思想。
考虑优化的话，发现枚举行或列中至少需要一步，因为这个题只能预处理优化，如果都不枚举，就相当于盲人摸象，无法预处理来优化。
因此要搜索枚举行或列，然后预处理并在列或行上跑DP，这样就可以少些枚举时间，多些转移时间。
这里我用了枚举行，预处理关于列上的数组，并在列上跑DP。
预处理和DP数组如下：
\(dp[i][j]\)表示前\(i\)列，已用\(j\)列得到的最小价值。\(ver[i]\)表示对于\(i\)列的上下绝对值差的和，\(del[i][j]\)表示\(i\)列和\(j\)列左右差的和
状态转移方程：
$dp[i][j] = min(dp[i][j],dp[i-k][j-1]+ver[i]+del[i-k][i]); $
50pts:
#include#include #include #include using namespace std;int n, m, r, c, dp1[109], dp2[109], a[19][19];int ans = 0x7fffffff;void Dp(){ int now = 0; for (int i = 1; i <= r; i++) for (int j = 2; j <= c; j++) now += abs(a[dp1[i]][dp2[j]] - a[dp1[i]][dp2[j-1]]); for (int i = 2; i <= r; i++) for (int j = 1; j <= c; j++) now += abs(a[dp1[i]][dp2[j]] - a[dp1[i-1]][dp2[j]]); ans = min(ans, now);}void dfs(int x, int y, int nr, int nc){ if (nc == c + 1) return Dp(); if ((x > n && nr != r + 1) || (y > m && nc != c + 1)) return; if (nr == r + 1) {//先枚举行，再枚举列 for (int i = y; i <= m; i++) { dp2[nc] = i; dfs(x, i + 1, nr, nc + 1); } return; } else for (int i = x; i <= n; i++) { dp1[nr] = i; dfs(i + 1, y, nr + 1, nc); }}int main(){ scanf("%d%d%d%d", &n, &m, &r, &c); for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) scanf("%d", &a[i][j]); dfs(1, 1, 1, 1); printf("%d\n", ans); return 0;}
100pts:
#include#include #include #include #include using namespace std;int n, m, r, c, ans = 2147483647;int a[19][19], hang[19], dp[19][19];int ver[19], del[19][19];//ver[i]表示对于i列的上下绝对值差的和，del[i][j]表示i列和j列左右差的和 inline void Dp(){ memset(dp, 123, sizeof(dp)); memset(ver, 0, sizeof(ver)); memset(del, 0, sizeof(del)); for (int i = 1; i <= m; i++)//枚举每一列i for (int j = 2; j <= r; j++) ver[i] += abs(a[hang[j]][i] - a[hang[j - 1]][i]) ; for (int i = 1; i <= m; i++) for (int k = i + 1; k <= m; k++) for (int j = 1; j <= r; j++) del[i][k] += abs(a[hang[j]][k] - a[hang[j]][i]); for (int i = 1; i <= m; i++) dp[i][1] = ver[i]; for (int i = 1; i <= m; i++) for (int j = 1; j <= c; j++) for (int k = 1; k < i && i - k >= j - 1; k++) dp[i][j] = min(dp[i][j], dp[i - k][j - 1] + ver[i] + del[i - k][i]); for (int i = c; i <= m; i++)//从c行后取最小值 ans = min(ans, dp[i][c]); } void dfs(int now, int pos){ if (now == r + 1) { Dp(); return; } if (pos == n + 1) return; for (int i = pos; i <= n; i++) hang[now] = i, dfs(now + 1, i + 1);}int main(){ scanf("%d%d%d%d", &n, &m, &r, &c); for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) scanf("%d", &a[i][j]); dfs(1, 1); printf("%d", ans);}