POJ 1160 Post Office (DP)

思路：dp[i][j]，表示放了i个邮局，满足了1 - j个村庄，那么现在多一个邮局，去满足一个区间的村庄，最好的位置肯定是放在中位数的位置，所以可以先预处理出前缀和，然后去转移dp[i][j] = min{dp[i - 1][k] + sum{j, k});

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 305;
const int INF = 0x3f3f3f3f;

int v, p, post
, dp

, sum
;

int main() {
while (~scanf("%d%d", &v, &p)) {
for (int i = 1; i <= v; i++) {
scanf("%d", &post[i]);
sum[i] = sum[i - 1] + post[i];
}
for (int i = 1; i <= v; i++) dp[0][i] = INF;
for (int i = 1; i <= p; i++) {
for (int j = 1; j <= v; j++) {
dp[i][j] = INF;
for (int k = 1; k <= j; k++) {
int mid = (k + j) / 2;
int pre = (mid - k + 1) * post[mid] - (sum[mid] - sum[k - 1]);
int suf = (sum[j] - sum[mid]) - (j - mid) * post[mid];
dp[i][j] = min(dp[i][j], dp[i - 1][k - 1] + pre + suf);
}
}
}
printf("%d\n", dp[p][v]);
}
return 0;
}