POJ - 1160 Post Office

POJ - 1160 

思路：

平行四边形不等式优化dp

dp[i][j]：前j个选i个作为邮局的最小答案

w[i][j]:i到j之间选一个作为邮局的最小距离和，肯定是选中间的

dp[i][j] = min{dp[i-1][k] + w[k+1][j]}

这个方程和石子归并类似，满足四边形不等式（一般打表找规律，不推不等式），所以决策变量满足决策单调性

假设s[i][j]为dp[i][j]的最优决策变量k,则s[i-1][j] <= s[i][j] <= s[i][j+1]

所以j这一个维度从大到小

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
#define LD long double
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<int, pii>
#define pdd pair<long double, long double>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head

const int N = 305;
int x[N], n, p;
int dp[N][N], s[N][N], sum[N];
inline int cal(int l, int r) {
    int m = l+r >> 1;
    return (m-l+1)*x[m] - (sum[m]-sum[l-1]) + sum[r] - sum[m] - (r-m)*x[m];
}
int main() {
    scanf("%d %d", &n, &p);
    for (int i = 1; i <= n; ++i) scanf("%d", &x[i]);
    for (int i = 1; i <= n; ++i) sum[i] = sum[i-1]+x[i];
    mem(dp, 0x3f);
    dp[0][0] = 0;

    for (int i = 1; i <= p; ++i) {
        s[i][n+1] = n-1;
        for (int j = n; j >= 1; --j) {
            for (int k = s[i-1][j]; k <= s[i][j+1]; ++k) {
                if(k+1 <= j && dp[i-1][k]+cal(k+1, j) < dp[i][j]) {
                    dp[i][j] = dp[i-1][k]+cal(k+1, j);
                    s[i][j] = k;
                }
            }
        }
    }
    printf("%d\n", dp[p][n]);
    return 0;
}

附打表代码及结果