HDU

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题目：有n个数，要分成若干组，每组至少L个数，每组的价值定义为每个数减去该组数的最小值的和，求各组数的价值和的最小值

思路：

先对序列arr按从小到大排序，求出前缀和

设dp[i]表示前i个数的合法分组的最小值

dp[i]=min{dp[j]+sum[i]-sum[j]-arr[j+1]*(i-j)}，j<=i-L

然后就是斜率优化了

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3f//0x3f3f3f3fconst int maxn=4e5+50;int q[maxn];LL dp[maxn];LL arr[maxn];LL sum[maxn];int head,tail,n,L;LL getdp(int i,int j){    return dp[j]+(sum[i]-sum[j])-arr[j+1]*(i-j);}LL gety(int k,int j){//yj-yk    return dp[j]-sum[j]+arr[j+1]*j-(dp[k]-sum[k]+arr[k+1]*k);}LL getx(int k,int j){//xj-xk    return arr[j+1]-arr[k+1];}int main(){//    freopen("D:\\input.txt","r",stdin);//    freopen("D:\\output.txt","w",stdout);    while(~scanf("%d%d",&n,&L)){        for(int i=1;i<=n;i++)            scanf("%I64d",&arr[i]);        sort(arr+1,arr+n+1);        sum[0]=dp[0]=0;        for(int i=1;i<=n;i++)            sum[i]=sum[i-1]+arr[i];        head=0;        tail=-1;        q[++tail]=0;        for(int i=1;i<=n;i++){            while(head<tail&&gety(q[head],q[head+1])<=(LL)i*getx(q[head],q[head+1]))                head++;            dp[i]=getdp(i,q[head]);            int j=i-L+1;            if(i-L+1<L) continue;//注意这里            while(head<tail&&gety(q[tail-1],q[tail])*getx(q[tail],j)>=gety(q[tail],j)*getx(q[tail-1],q[tail]))                tail--;            q[++tail]=j;        }        printf("%I64d\n",dp[n]);    }    return 0;}

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