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hdu 3480 Division (斜率优化)

时间:2014-07-02 08:52:54      阅读:133      评论:0      收藏:0      [点我收藏+]

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Division

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 999999/400000 K (Java/Others)
Total Submission(s): 2676    Accepted Submission(s): 1056


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

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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
 

题意:
将含有N个元素的一个集合分成M个子集,使得每个子集的最大值与最小值平方差的和最小。

思路:
可以想到贪心将元素从小到大排序,很快可以得出dp[i][j]-前i个子集分j个元素的最小花费,
dp方程  dp[i][j]=dp[i-1][k]+(a[j]-a[k+1]);但是需要三重循环,时间复杂度接受不了,所以需要优化,这题斜率优化和四边形不等式都可以,我采用的是刚学习的斜率优化。
设k1<k2<j,如果k1比k2优,则有dp[i-1][k1]+(a[j]-a[k1+1])^2<dp[i-1][k2]+(a[j]-a[k2+1])^2;
化简得 ( dp[i-1][k1]+a[k1+1]^2-(dp[i-1][k2]+a[k2+1]^2) )/(a[k1+1]-a[k2+1])<=2a[j];
类比斜率,可以得出 
y:dp[i-1][k1]+a[k1+1]^2;
x:   a[k1+1];
于是和之前见到的斜率优化有了一个相同的式子 (y1-y2)/(x1-x2)<k;于是可以用一个单调队列来维护一个下凸函数优化dp了。
斜率优化可以用滚动数组实现,节约内存一些。

代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define maxn 205
#define MAXN 100005
#define mod 100000000
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

int n,m,ans,cnt,tot,flag;
int a[10005],dp[2][10005];
int q[10005];

int sqr(int x)
{
    return x*x;
}
int get(int i,int k)
{
    return dp[i][k]+a[k+1]*a[k+1];
}
void solve()
{
    int i,j,t,dx1,dy1,dx2,dy2,k1,k2,turn=0;
    int head,tail;
    sort(a+1,a+n+1);
    memset(dp,0x3f,sizeof(dp));
    dp[0][0]=0;
    for(i=1; i<=m; i++)
    {
        head=0; tail=-1;
        q[++tail]=i-1;
        for(j=i; j<=n; j++)
        {
            while(head<tail) // 将不再更新答案的点删除
            {
                k1=q[head+1];
                k2=q[head];
                dx1=a[k1+1]-a[k2+1];
                dy1=get(turn,k1)-get(turn,k2);
                if(dy1<=dx1*2*a[j]) head++;
                else break ;
            }
            dp[turn^1][j]=dp[turn][q[head]]+sqr(a[j]-a[q[head]+1]);
            while(head<tail) // 维护下凸曲线
            {
                dx1=a[j+1]-a[q[tail]+1];
                dy1=get(turn,j)-get(turn,q[tail]);
                dx2=a[q[tail]+1]-a[q[tail-1]+1];
                dy2=get(turn,q[tail])-get(turn,q[tail-1]);
                if(dy1*dx2<=dy2*dx1) tail--;
                else break ;
            }
            q[++tail]=j;
        }
        turn^=1;
    }
    ans=dp[turn][n];
}
int main()
{
    int i,j,t,test=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
        }
        solve();
        printf("Case %d: %d\n",++test,ans);
    }
    return 0;
}





hdu 3480 Division (斜率优化),布布扣,bubuko.com

hdu 3480 Division (斜率优化)

标签:des   style   http   java   color   strong   

原文地址:http://blog.csdn.net/tobewhatyouwanttobe/article/details/36209591

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