HDU 3045 (斜率优化)

题目链接：点击这里

题意：n个数，分成若干块，每一块的数量不能小于k，每一块的花费是所有的数减去块中最小值的和。每一块和的最小值。

显然要排序以后分块，设fi表示(1,i)分块后的最小和，那么有fi=min{fj+sumi−sumj−aj+1∗(i−j)∥∥i−j≥k}

假设j比k更优(k≤j)fj+sumi−sumj−aj+1×(i−j)≤fk+sumi−sumk−ak+1×(i−k)

整理一下就是斜率式子了fj+j×aj+1−sumj−fk−k×ak+1+sumkaj+1−ak+1≤i

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
#define maxn 400005

long long dp[maxn];
int n, k, que[maxn];
long long a[maxn], sum[maxn];

long long scan () {
char ch=' ';
while(ch<'0'||ch>'9')ch=getchar();
long long x=0;
while(ch<='9'&&ch>='0')x=x*10+ch-'0',ch=getchar();
return x;
}

long long up (int i, int j) {
return dp[i]+1LL*i*a[i+1]-sum[i] - (dp[j]+1LL*j*a[j+1]-sum[j]);
}

long long down (int i, int j) {
return a[i+1]-a[j+1];
}

int main () {
while (scanf ("%d%d", &n, &k) == 2) {
sum[0] = 0;
for (int i = 1; i <= n; i++) a[i] = scan ();;
sort (a+1, a+1+n);
for (int i = 1; i <= n; i++) sum[i] = sum[i-1]+a[i];

dp[0] = 0;
for (int i = 1; i < k*2 && i <= n; i++) {
dp[i] = sum[i] - 1LL*i*a[1];
}
int L = 0, R = 0;
que[R++] = 0;
que[R++] = k;
for (int i = 2*k; i <= n; i++) {
while (L+1 < R && up (que[L+1], que[L]) <= 1LL*i*down (que[L+1], que[L]))
L++;
int j = que[L];
dp[i] = dp[j] + sum[i]-sum[j]-a[j+1]*(i-j);
while (L+1 < R && up (i-k+1, que[R-1])*down (que[R-1], que[R-2]) <=
up (que[R-1], que[R-2])*down (i-k+1, que[R-1]))
R--;
que[R++] = i-k+1;
}
printf ("%lld\n", dp
);
}
return 0;
}