POJ1160 [IOI2000]邮局 [四边形不等式优化dp]

$邮局$

题目描述见链接 .

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$\color{red}{正解部分}$

设 $F[i, j]$ 表示前 $i$ 个村庄, 建立 $j$ 个邮局的最优值,

$状态转移$: $F[i, j] = \min_{0 \le k < i}(F[k, j-1] + w(k+1, i))$,

其中 $w(l, r)$ 表示在 $[l,r]$ 放置一个邮局, 管辖 $[l,r]$ 的村庄的最小值, 显然放在 中位数 位置最优,现在考虑如何 $O(1)$ 计算 $w(l, r)$,

设 $sum\_l[i]$ 表示 $[1, i]$ 的距离前缀和, $sum\_r[i]$ 表示 $[i, N]$ 的距离后缀和, 可以 $O(N)$ 预处理 .

设 $m$ 为 $[l, r]$ 的中位数所在村庄编号,

则 $w(l, r) = (m-l)\times A[m] - (sum\_l[m-1] - sum\_l[l-1]) + (sum\_r[m+1]-sum\_r[r+1]) - (r-m)\times A[m]$

然后经过 打表 可知, $F$ 满足四边形不等式: $F[i+1,j+1]+F[i,j] \le F[i+1, j] + F[i, j+1]$

同时与此对应的最优决策 $p[i,j]$ 也 满足决策单调性: $p[i,j-1] \le p[i, j] \le p[i+1, j]$.

于是可以使用 四边形不等式 优化 $O(N^2K) \rightarrow O(NK)$ ,

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$\color{red}{实现部分}$

先给出 打表/暴力 程序

#include<bits/stdc++.h>
#define reg register
typedef long long ll;

const int maxn = 3005;

int read(){
        char c;
        int s = 0, flag = 1;
        while((c=getchar()) && !isdigit(c))
                if(c == '-'){ flag = -1, c = getchar(); break ; }
        while(isdigit(c)) s = s*10 + c-'0', c = getchar();
        return s * flag;
}

int N;
int K;
int A[maxn];
int p[maxn][maxn];

ll sl[maxn];
ll sr[maxn];
ll sum_l[maxn];
ll sum_r[maxn];
ll F[maxn][maxn];
ll w[maxn][maxn];

int main(){
        freopen("a.in", "r", stdin);
        freopen("a.out", "w", stdout);
        N = read(); K = read();
        for(reg int i = 1; i <= N; i ++) A[i] = read();
        std::sort(A+1, A+N+1);
        for(reg int i = 1; i <= N; i ++) sum_l[i] = sum_l[i-1] + A[i], sl[i] = sum_l[i] + sl[i-1];
        for(reg int i = N; i >= 1; i --) sum_r[i] = sum_r[i+1] + A[i], sr[i] = sum_r[i] + sr[i+1];
        for(reg int i = 1; i <= N; i ++)
                for(reg int j = i; j <= N; j ++){
                        int m = i+j >> 1;
                        w[i][j] = (1ll*m-i)*A[m] - (sum_l[m-1] - sum_l[i-1]) + (sum_r[m+1] - sum_r[j+1]) - (1ll*j-m)*A[m];
                } 
        memset(F, 0x3f, sizeof F); F[0][0] = 0;
        for(reg int i = 1; i <= N; i ++)
                for(reg int j = 1; j <= i; j ++){
                        for(reg int k = 0; k < i; k ++){
                                ll t = F[k][j-1] + w[k+1][i];
                                if(t < F[i][j]){
                                        F[i][j] = t;
                                        p[i][j] = k;
                                }
                        }
                }
        for(reg int i = 1; i <= N; i ++)
                for(reg int j = 1; j <= i; j ++)
                        if(i!=N && j != 1 && (p[i][j-1] > p[i][j] || p[i][j] > p[i+1][j])){
                                printf("False\n");
                        }else if(i != N && j != i && (F[i+1][j+1] + F[i][j] > F[i+1][j] + F[i][j+1])){
                                printf("False\n");
                        }
        printf("%lld\n", F[N][K]);
        return 0;
}

---

$Accept$ $code$

#include<cstdio>
#include<cctype>
#include<cstring>
#include<algorithm>
#define reg register
typedef long long ll;

const int maxn = 3005;

int read(){
        char c;
        int s = 0, flag = 1;
        while((c=getchar()) && !isdigit(c))
                if(c == '-'){ flag = -1, c = getchar(); break ; }
        while(isdigit(c)) s = s*10 + c-'0', c = getchar();
        return s * flag;
}

int N;
int K;
int A[maxn];
int p[maxn][305];

ll sum_l[maxn];
ll sum_r[maxn];
ll F[maxn][305];
ll w[maxn][maxn];

int main(){
        N = read(); K = read();
        for(reg int i = 1; i <= N; i ++) A[i] = read();
        std::sort(A+1, A+N+1);
        for(reg int i = 1; i <= N; i ++) sum_l[i] = sum_l[i-1] + A[i];
        for(reg int i = N; i >= 1; i --) sum_r[i] = sum_r[i+1] + A[i];
        for(reg int i = 1; i <= N; i ++)
                for(reg int j = i; j <= N; j ++){
                        int m = i+j >> 1;
                        w[i][j] = (1ll*m-i)*A[m] - (sum_l[m-1] - sum_l[i-1]) + (sum_r[m+1] - sum_r[j+1]) - (1ll*j-m)*A[m];
                } 

        memset(F, 0x3f, sizeof F); F[0][0] = 0;

        for(reg int j = 1; j <= K; j ++){
                p[N+1][j] = N-1;
                for(reg int i = N; i >= 1; i --)
                        for(reg int k = p[i][j-1]; k <= p[i+1][j]; k ++){ 
                                ll t = F[k][j-1] + w[k+1][i];
                                if(t < F[i][j]) F[i][j] = t, p[i][j] = k; 
                        }
        }
        printf("%lld\n", F[N][K]);
        return 0;
}