Division
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 999999/400000 K (Java/Others)
Total Submission(s): 5344 Accepted Submission(s):
2115
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.
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.
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
2
3 2
1 2 4
4 2
4 7 10 1
Sample Output
Case 1: 1
Case 2: 18
Hint
The answer will fit into a 32-bit signed integer.
Source
Recommend
四边形不等式好恶心。。
首先对所有的数据排序（根据方差的性质贪心）
我们用$dp[i][j]$表示前$j$个数，分为$i$段的最小代价
朴素的转移的话枚举前一段的断点
然后根据……&*（）￥#%……&我们可以知道这玩意儿满足四边形不等式
然后愉快的套上板子就好啦
#include<cstdio> #include<cstring> #include<algorithm> const int MAXN=10001,INF=1e9+10; using namespace std; inline int read() { char c=getchar();int x=0,f=1; while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} return x*f; } int dp[MAXN][MAXN],s[MAXN][MAXN],a[MAXN]; int mul(int x){return x*x;} int main() { int Test=read(),cnt=0; while(Test--) { int N=read(),M=read(); for(int i=1;i<=N;i++) a[i]=read();sort(a+1,a+N+1); for(int i=1;i<=N;i++) dp[1][i]=mul(a[i]-a[1]),s[1][i]=1; for(int i=2;i<=M;i++) { s[i][N+1]=N-1;//边界 for(int j=N;j>=i;j--) { int mn=INF,mnpos=-1; for(int k=s[i-1][j];k<=s[i][j+1];k++) { if(dp[i-1][k]+mul(a[j]-a[k+1])<mn) { mn=dp[i-1][k]+mul(a[j]-a[k+1]); mnpos=k; } } dp[i][j]=mn; s[i][j]=mnpos; } } printf("Case %d: %d\n",++cnt,dp[M][N]); } return 0; }
