HDU 3480 Division(斜率优化DP)
Division
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
![](http://acm.hdu.edu.cn/data/images/C295-1003-1.jpg)
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
![](http://acm.hdu.edu.cn/data/images/C295-1003-1.jpg)
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
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<k<j;
利用斜率优化DP,整理下就可以出来了。
#include<stdio.h> #include<string.h> #include<algorithm> #include<iostream> 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<stdio.h> #include<iostream> #include<string.h> #include<algorithm> 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; }
人一我百!人十我万!永不放弃~~~怀着自信的心,去追逐梦想