斜率優化dp

hdu3045

在做了幾個入門的斜率dp題之後寫這個題,其實這個題的方程跟之前的入門題hdu3507差不多 ,

dp[i] = min(dp[i] , dp[j] + (sum[i] - sum[j]) - (i-j) *(num[i+1]) ) ; {0 <= j <= i - T}
可以看出來其實基本上沒什麼不同,唯一有區別的是j的取值範圍.0到i-T
這是我一開始寫的版本

#include<cmath>
#include<algorithm>
#include<cstring>
#include<string>
#include<set>
#include<map>
#include<time.h>
#include<cstdio>
#include<vector>
#include<list>
#include<stack>
#include<queue>
#include<iostream>
#include<stdlib.h>
using namespace std;
#define  LONG long long
const int   INF=0x3f3f3f3f;
const LONG  MOD=1e9+ 7;
const double PI=acos(-1.0);
#define clrI(x) memset(x,-1,sizeof(x))
#define clr0(x) memset(x,0,sizeof x)
#define clr1(x) memset(x,INF,sizeof x)
#define clr2(x) memset(x,-INF,sizeof x)
#define EPS 1e-10
#define lson  l , mid , rt<< 1
#define rson  mid + 1 ,r , (rt<<1)+1
#define root 1, m , 1
LONG num[400200] ;
LONG dp[450010] ;
LONG que[410000] ;
LONG Sum[410000] ;
LONG Up(LONG  j, LONG k)
{
    return (dp[j] - dp[k]) - (Sum[j] - Sum[k]) + j * num[j+1] - k*num[k+1] ;
}
LONG Down(LONG j , LONG k)
{
    return num[j+1] - num[k+1] ;
}
int main()
{
//freopen("/home/weyoung/桌面/inn","r",stdin);

    int T;
    int n ;
    while( cin>> n >>T)
    {
        clr1(dp) ;
        dp[0] = 0 ;
        Sum[0] = 0 ;
        for(int i =1; i<= n ; ++i)
            scanf("%lld",&num[i]) ;
        sort(num+1 , num + n + 1  ) ;
        for(int i = 1;i<=n ; ++i)
        Sum[i] = Sum[i - 1] + num[i] ;
        int head =  1, tail = 1;
        que[1] = 0 ;
        for(int i = 1; i<= n ; ++i)
        {
            if(i < T ) continue ;
            while(1)
            {
                if(head >= tail )break ;
                if( que[head+1] <= i-T && Up(que[head+1] , que[head] )<= i * Down(que[head+1], que[head]))
                head ++ ;
                else break ;
            }
            int t = que[head] ;
            dp[i] = dp[t] + (Sum[i] - Sum[t]) - (i - t) *num[t+1] ;
            while(1)
            {
                if(head >= tail) break ;
                if(Up(i,que[tail]) * Down(que[tail] , que[tail-1]) <= Up(que[tail] , que[tail-1]) * Down(i ,que[tail]))
                    tail -- ;
                else
                    break ;
            }
            que[++tail] = i ;
        }
//        for(int i =1; i<= n ; ++i)
//            printf("%lld ",dp[i]); cout<<endl;
        cout<<dp[n]<<endl;
    }
}

這樣寫看起來好像也沒什麼地方不對,但是有個重要的問題,當我們在往隊列中插入元素的時候,要滿足g[i,j] > g[j , k] ,而這個題目j的取值範圍是i-T,但是這份代碼裏,實際上往隊列裏插的是i,這就導致了當前隊列中有些元素不滿足g[i,j] > g[j.k]但是可以滿足g[i-T,j] > g[j,k] ,把有些有效元素拋出瞭解集,最後導致轉移不充分,結果偏大.
正解應該是這樣

#include<cmath>
#include<algorithm>
#include<cstring>
#include<string>
#include<set>
#include<map>
#include<time.h>
#include<cstdio>
#include<vector>
#include<list>
#include<stack>
#include<queue>
#include<iostream>
#include<stdlib.h>
using namespace std;
#define  LONG long long
const LONG   INF=0x3f3f3f3f;
const LONG  MOD=1e9+ 7;
const double PI=acos(-1.0);
#define clrI(x) memset(x,-1,sizeof(x))
#define clr0(x) memset(x,0,sizeof x)
#define clr1(x) memset(x,INF,sizeof x)
#define clr2(x) memset(x,-INF,sizeof x)
#define EPS 1e-10
#define lson  l , mid , rt<< 1
#define rson  mid + 1 ,r , (rt<<1)+1
#define root 1, m , 1
LONG num[400200] ;
LONG dp[450010] ;
LONG que[410000] ;
LONG Sum[410000] ;
LONG Up(LONG  j, LONG k)
{
    return (dp[j] - dp[k]) - (Sum[j] - Sum[k]) + j * num[j+1] - k*num[k+1] ;
}
LONG Down(LONG j , LONG k)
{
    return num[j+1] - num[k+1] ;
}
int main()
{


    LONG n , T ;
    while( cin>> n >>T)
    {
        dp[0] = 0 ;
        Sum[0] = 0 ;
        for(LONG i =1; i<= n ; ++i)
            scanf("%lld",&num[i]) ;
        sort(num + 1 , num + n + 1  ) ;
        for(LONG i = 1;i<=n ; ++i)
        Sum[i] = Sum[i - 1] + num[i] ;
        LONG head = 1, tail = 1;
        que[1] = 0 ;
        for(LONG i = T; i<= n ; ++i)
        {
            LONG judge = 0;
            while(1)
            {
                if(head >= tail )break ;
                if(  Up(que[head+1] , que[head] )<= i * Down(que[head+1], que[head]))
                head ++ ;
                else break ;
            }
            LONG t = que[head] ;

            dp[i] = dp[t] + (Sum[i] - Sum[t]) - (i - t) *num[t+1] ;
            int p = i - T + 1;
            if(p < T) continue; // 注意這裏
            while(1)
            {
                if(head >= tail) break ;
                if(Up(p,que[tail]) * Down(que[tail] , que[tail-1]) <= Up(que[tail] , que[tail-1]) * Down(p ,que[tail]))
                    tail -- ;
                else
                    break ;
            }
            que[++tail] = p;
        }
        cout<<dp[n]<<endl;
    }
}
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