dp斜率优化 Hdu 3480 Division 题解

累加器传送门：

：http://blog.csdn.net/NOIAu/article/details/71775000

题目传送门：

https://vjudge.net/problem/HDU-3480

题目：

Little D is really interested in the theorem of sets recently. There’s a problem that confused him a long time.

Let T be a set of integers. Let the MIN be the minimum integer in T and MAX be the maximum, then the cost of set T if defined as (MAX – MIN)^2. Now given an integer set S, we want to find out M subsets S1, S2, …, SM of S, such that



and the total cost of each subset is minimal.

输入：

The input contains multiple test cases.

In the first line of the input there’s an integer T which is the number of test cases. Then the description of T test cases will be given.

For any test case, the first line contains two integers N (≤ 10,000) and M (≤ 5,000). N is the number of elements in S (may be duplicated). M is the number of subsets that we want to get. In the next line, there will be N integers giving set S.

输出：

For each test case, output one line containing exactly one integer, the minimal total cost. Take a look at the sample output for format.

样例输入：

2

3 2

1 2 4

4 2

4 7 10 1

样例输出：

Case 1: 1

Case 2: 18

先排序，然后

用dp[i][j]表示前j个数，分为i段的最优值

容易写出：

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

由于数据已经有序，自然cost[k+1][j]的值应该等于（a[j]-a[k+1]）^2

所以写出转移方程

dp[i][j]=dp[i-1][k]+a[k+1]^2-2 *a[j] *a[k+1]+a[j]^2

令b=dp[i][j]

y=dp[i-1][k]+a[k+1]^2

k=2*a[j]

x=a[k+1]

二维的斜率优化就可以了

代码如下

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define MAXN 10000+10

using namespace std;

int cn;
int a[MAXN];
int q[MAXN];
int head,tail;
int n,m;
int dp[MAXN][MAXN];
int T;

void init(){
scanf("%d%d",&n,&m);
for(register int i=1;i<=n;i++) scanf("%d",&a[i]);
sort(a+1,a+n+1);
}

void dpp(){
for(int i=1;i<=n;i++) dp[1][i]=(a[i]-a[1])*(a[i]-a[1]);
for(int i=2;i<=m;i++){
head=tail=0;
q[tail++]=i-1;
for(int j=i;j<=n;j++){
while(head+1<tail){
int y1=dp[i-1][q[head]]+a[q[head]+1]*a[q[head]+1];
int y2=dp[i-1][q[head+1]]+a[q[head+1]+1]*a[q[head+1]+1];
int x1=a[q[head]+1];
int x2=a[q[head+1]+1];
if(y1-2*a[j]*x1>=y2-2*a[j]*x2)head++;
else break;
}
int k=q[head];
dp[i][j]=dp[i-1][k]+(a[j]-a[k+1])*(a[j]-a[k+1]);
while(head+1<tail){
int x1=a[j+1]-a[q[tail-1]+1];
int y1=dp[i-1][j]+a[j+1]*a[j+1]-(dp[i-1][q[tail-1]]+a[q[tail-1]+1]*a[q[tail-1]+1]);
int x2=a[q[tail-1]+1]-a[q[tail-2]+1];
int y2=dp[i-1][q[tail-1]]+a[q[tail-1]+1]*a[q[tail-1]+1]-(dp[i-1][q[tail-2]]+a[q[tail-2]+1]*a[q[tail-2]+1]);
if(x1*y2-x2*y1>=0)tail--;
else break;
}
q[tail++]=j;
}
}
printf("Case %d: %d\n",++cn,dp[m]
);
}

int main(){
scanf("%d",&T);
while(T--){
init();
dpp();
}
}