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 400005long 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[n]);    }    return 0;}