一道经典的DP题目，应该并没有提高＋拉莫难。。。。。。。。。。。。。。。。。。。。。
思路：
搞懂题意后(这点也很重要),我们首先考虑棋盘的分割情况，不难发现：
然后我们考虑状态的设计和转移：
AC代码吧:
#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> using namespace std; int read() { int x = 0; int k = 1; char c = getchar(); while (c > '9' || c < '0') if (c == '-') c = getchar(), k = -1; else c = getchar(); while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar(); return k * x; } int n, b, x; int a[10][10]; int s[10][10]; int dp[10][10][10][10][17]; int calc(int x1, int y1, int x2, int y2) { return s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]; } int main() { n = read(); for (int i = 1; i <= 8; ++i) for (int j = 1; j <= 8; ++j) { if (j == 1) b = 0; a[i][j] = read(); b += a[i][j]; s[i][j] = s[i - 1][j] + b; } for (int i = 1; i <= 8; ++i) for (int j = 1; j <= 8; ++j) for (int o = i; o <= 8; ++o) for (int k = j; k <= 8; ++k) dp[i][j][o][k][0] = calc(i, j, o, k), dp[i][j][o][k][0] *= dp[i][j][o][k][0]; for (int p = 1; p < n; ++p) for (int i = 1; i <= 8; ++i) for (int j = 1; j <= 8; ++j) for (int o = i; o <= 8; ++o) for (int k = j; k <= 8; ++k) { int minn = (1 << 30) + 5555555; for (int l = j; l < k; ++l) minn = min(minn, min(dp[i][l + 1][o][k][0] + dp[i][j][o][l][p - 1], dp[i][l + 1][o][k][p - 1] + dp[i][j][o][l][0])); for (int l = i; l < o; ++l) minn = min(minn, min(dp[l + 1][j][o][k][0] + dp[i][j][l][k][p - 1], dp[l + 1][j][o][k][p - 1] + dp[i][j][l][k][0])); dp[i][j][o][k][p] = minn; } printf("%d", dp[1][1][8][8][n - 1]); }