标签：oid lan bool 预处理 turn 方案 cst string name
思路：分析公式，我们可以发现平均值那一项和我们怎么分的具体方案无关，影响答案的是每个矩阵的矩阵和的平方，由于数据很小，我们可以预处理出每个矩阵的和的平方，执行状态转移。
设dp[l1][r1][l2][r2][k]是矩阵l1,r1,l2,r2切割k次的最小值，我们可以枚举是横着切还是竖着切执行状态转移。
代码：
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define INF 0x3f3f3f3f
using namespace std;
int dp[10][10][10][10][16], sum[10][10][10][10];
bool v[10][10][10][10][16];
int a[10][10];
void init() {
for (int l1 = 1; l1 <= 8; l1++)
for (int r1 = 1; r1 <= 8; r1++)
for (int l2 = l1; l2 <= 8; l2++)
for (int r2 = r1; r2 <= 8; r2++) {
int ans = 0;
for (int i = l1; i <= l2; i++)
for (int j = r1; j <= r2; j++) {
ans += a[i][j];
}
sum[l1][r1][l2][r2] = ans * ans;
}
}
int solve(int l1, int r1, int l2, int r2, int k) {
if (v[l1][r1][l2][r2][k]) return dp[l1][r1][l2][r2][k];
if (k == 1) {
return dp[l1][r1][l2][r2][k] = sum[l1][r1][l2][r2];
}
int ans = INF;
for (int i = l1; i < l2; i++) {
ans = min(ans, min(solve(l1, r1, i, r2, k - 1) + sum[i + 1][r1][l2][r2], solve(i + 1, r1, l2, r2, k - 1) + sum[l1][r1][i][r2]));
}
for (int i = r1; i < r2; i++) {
ans = min(ans, min(solve(l1, r1, l2, i, k - 1) + sum[l1][i + 1][l2][r2], solve(l1, i + 1, l2, r2, k - 1) + sum[l1][r1][l2][i]));
}
v[l1][r1][l2][r2][k] = 1;
return dp[l1][r1][l2][r2][k] = ans;
}
int main() {
int n, tot = 0;
scanf("%d", &n);
for (int i = 1; i <= 8; i ++) {
for (int j = 1; j <= 8 ; j++) {
scanf("%d", &a[i][j]);
tot += a[i][j];
}
}
init();
solve(1, 1, 8, 8, n);
double ans = sqrt(dp[1][1][8][8][n] * 1.0 / n - (tot * 1.0 / n) * (tot * 1.0 / n));
printf("%.3f\n", ans);
}
标签：oid lan bool 预处理 turn 方案 cst string name
原文地址：https://www.cnblogs.com/pkgunboat/p/10339802.html
