HDU 3480 Division（斜率优化+二维DP）

　　　　　　　　　　　　　　　　Division

　　　　　　　　　　　　　　　　　　　　　　Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 999999/400000 K (Java/Others)
　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Total Submission(s): 3984 Accepted Submission(s): 1527


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For
each test case, output one line containing exactly one integer, the
minimal total cost. Take a look at the sample output for format.

[align=left]Sample Input[/align]

2
3 2
1 2 4
4 2
4 7 10 1

[align=left]Sample Output[/align]

Case 1: 1
Case 2: 18

Hint

The answer will fit into a 32-bit signed integer.

[align=left]Source[/align]
2010 ACM-ICPC Multi-University Training Contest（5）——Host by BJTU

【思路】

斜率优化+分配式DP。

设f[i][j]表示将前i个分作j个集合所得最小消费，则有转移方程式：

f[i][j]=min{ f[k][j-1]+(A[k]-A[j+1])^2 }

若有k>l，且决策k优于决策l则有：

f[k][j-1]-f[l][j-1]+sq(A[k+1])-sq(A[l+1]) <= 2*(A[k+1]-A[l+1])*A[i]

先进行j循环枚举f[][j]，每一层维护一个单调队列即可。

乘除耗费时间悬殊，如果直接除这个题就超时了。

【代码】

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;

typedef double Do;
const int N = 1e4+10;
const int M = 5000+10;

int f
[M],A
,q
;
int n,m,L,R;
int sq(int x) { return x*x; }
int UP(int l,int k,int j) {
return f[k][j-1]-f[l][j-1]+sq(A[k+1])-sq(A[l+1]);
}
int DN(int l,int k,int j) {
return 2*(A[k+1]-A[l+1]);
}
void read(int& x) {
char c=getchar(); while(!isdigit(c)) c=getchar();
x=0; while(isdigit(c)) x=x*10+c-'0' , c=getchar();
}
int main() {
int T,kase=0;
read(T);
while(T--) {
read(n),read(m);
for(int i=1;i<=n;i++) read(A[i]);
sort(A+1,A+n+1);
for(int i=1;i<=n;i++) f[i][1]=sq(A[i]-A[1]);    //初始化第一层
for(int j=2;j<=m;j++) {
L=R=0;
for(int i=1;i<=n;i++) {
while(L<R && UP(q[L],q[L+1],j)<=A[i]*DN(q[L],q[L+1],j)) L++;
int t=q[L];
f[i][j]=f[t][j-1]+sq(A[i]-A[t+1]);
while(L<R && UP(q[R-1],q[R],j)*DN(q[R],i,j)>=UP(q[R],i,j)*DN(q[R-1],q[R],j)) R--;
q[++R]=i;
}
}
printf("Case %d: %d\n",++kase,f
[m]);
}
return 0;
}