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): 5883    Accepted Submission(s): 2337

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.

 

 

也是可以用斜率优化的一个dp

dp[i][j] 表示用j个数 划分 i个集合的最小花费

则 dp[i][j] = min{ dp[i-1][k] + (a[j]-a[k+1])^2 } (i-1<=k<j)

同理 假设 两个 j,k 使得 cost[j] <= cost[k]  得到一个斜率式子

然后用单调队列 优化一下 就可以了

#include <iostream>
#include <cstdio>
#include <string.h>
#include <algorithm>
using namespace std;

const int N = 1e4+10; const int M = 5e3+10;
#define rep(i,l,r) for(int i=l;i<=r;i++)
int n,m;
int a[N], dp[M][N], Q[N];

void init() {
    scanf("%d %d", &n, &m);
    rep(i,1,n) 
        scanf("%d",&a[i]);
    sort(a+1,a+1+n);
}

/* 朴素 o(m*n*n)
void solve() {
    memset(dp,0x3f,sizeof(dp));
    dp[0][0]=0;
    rep(i,1,m) {
        rep(j,i,n) {
            rep(k,i-1,j-1) {
                dp[i][j] = min(dp[i][j], dp[i-1][k] + (a[j]-a[k+1])*(a[j]-a[k+1]));
            }
        }
    }
    printf("%d
", dp[m][n]);
}
*/

void solve() {
    memset(dp,0x3f,sizeof(dp));
    dp[0][0] = 0;
    rep(i,1,m) {
        int st=0,ed=0;
        Q[ed++] = i-1;
        rep(j,i,n) {
            while (st+1 < ed) {
                int k1 = Q[st];
                int k2 = Q[st+1];
                int x1 = a[k1+1];
                int x2 = a[k2+1];
                int y1 = dp[i-1][k1] + a[k1+1]*a[k1+1];
                int y2 = dp[i-1][k2] + a[k2+1]*a[k2+1];
                if(y2 - y1 <= 2 * a[j] *  (x2 - x1)) st++;
                else break;
            }
            int k = Q[st];
            dp[i][j] = dp[i-1][k] + (a[j] - a[k+1]) * (a[j] - a[k+1]);
            while (st+1 < ed) {
                int k1 = Q[ed-2], k2 = Q[ed-1], k3 = j;
                int x1 = a[k1+1], x2 = a[k2+1], x3 = a[j+1];
                int y1 = dp[i-1][k1] + a[k1+1]*a[k1+1];
                int y2 = dp[i-1][k2] + a[k2+1]*a[k2+1];
                int y3 = dp[i-1][k3] + a[k3+1]*a[k3+1];
                if((y3-y2)*(x2-x1) <=(y2-y1)*(x3-x2)) ed--;
                else break;
            }
            Q[ed++] = j;
        }
    }
}

int main () {
    int T; scanf("%d",&T);
    rep(cas,1,T) {
        init();
        solve();
        printf("Case %d: %d
",cas,dp[m][n]);
    }
    return 0;
}