作者：管怡6440_368 | 来源：互联网 | 2023-08-30 07:46
TimeLimit:100005000MS(JavaOthers) MemoryLimit:999999400000K(JavaOthers)TotalSubmission(
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 999999/400000 K (Java/Others)
Total Submission(s): 1672 Accepted Submission(s): 630
Problem Description
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.
Input
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.
Output
For each test case, output one line containing exactly one integer, the minimal total cost. Take a look at the sample output for format.
Sample Input
Sample Output
Case 1: 1 Case 2: 18
Hint
The answer will fit into a 32-bit signed integer.
Source
2010 ACM-ICPC Multi-University Training Contest（5）——Host by BJTU
Recommend
zhengfeng
首先从小到大排序。
然后设 dp[i][j]表示前j个数分成i组的最小花费。
则 dp[i][j]=min{dp[i-1][k]+(a[j]-a[k+1])^2} 0
利用斜率优化DP，整理下就可以出来了。
#include
#include<string.h>
#include
#include
using namespace std;
const int MAXN=10010;
const int MAXM=5010;
int a[MAXN];
int dp[MAXM][MAXN];
int n,m;
int q[MAXN];
int head,tail;
int DP()
{
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 p1=q[head];
int p2=q[head+1];
int x1=a[p1+1];
int x2=a[p2+1];
int y1=dp[i-1][p1]+x1*x1;
int y2=dp[i-1][p2]+x2*x2;
if((y2-y1)<=2*a[j]*(x2-x1))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 p1=q[tail-2];
int p2=q[tail-1];
int p3=j;
int x1=a[p1+1];
int x2=a[p2+1];
int x3=a[p3+1];
int y1=dp[i-1][p1]+x1*x1;
int y2=dp[i-1][p2]+x2*x2;
int y3=dp[i-1][j]+x3*x3;
if((y3-y2)*(x2-x1)<=(y2-y1)*(x3-x2))tail--;
else break;
}
q[tail++]=j;
}
}
return dp[m][n];
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int T;
scanf("%d",&T);
int iCase=0;
while(T--)
{
iCase++;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
sort(a+1,a+n+1);
printf("Case %d: %d\n",iCase,DP());
}
return 0;
}
另外还有四边形不等式优化，还没有非常理解，正在学习中
/*
HDU 3840
C++ 2884ms C++
*/
#include
#include
#include<string.h>
#include
using namespace std;
const int MAXN=10010;
const int MAXM=5010;
int a[MAXN];
int s[MAXN][MAXM];
int dp[MAXN][MAXM];
int main()
{
int n,m;
int T;
scanf("%d",&T);
int iCase=0;
while(T--)
{
iCase++;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
sort(a+1,a+n+1);
for(int i=1;i<=n;i++)
{
dp[i][1]=(a[i]-a[1])*(a[i]-a[1]);
s[i][1]=1;
}
for(int k=2;k<=m;k++)
{
s[n+1][k]=n-1;
for(int i=n;i>=k;i--)
{
dp[i][k]=dp[k-1][k-1]+(a[i]-a[k])*(a[i]-a[k]);
s[i][k]=k;
for(int j=s[i][k-1];j<=s[i+1][k];j++)
{
int temp=dp[j][k-1]+(a[i]-a[j+1])*(a[i]-a[j+1]);
if(temp<dp[i][k])
{
dp[i][k]=temp;
s[i][k]=j;
}
}
}
}
printf("Case %d: %d\n",iCase,dp[n][m]);
}
return 0;
}
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