Division Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 999999/400000 K (Java/Others)Total Submission(s): 5344 Accepted Submission(s): 2115 Pro ...
Division
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 999999/400000 K (Java/Others)
Total Submission(s): 5344 Accepted Submission(s):
2115
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 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 2010 ACM-ICPC Multi-University Training Contest（5）——Host by BJTU
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#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;//邊界 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; }
