HDU 3415 Max Sum of Max-K-sub-sequence ( 单调队列 )

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题目链接~~>

做题感悟：看这题是就很有单调队列的赶脚，但是还是花费了很长时间做出来。

解题思路：

动态方程很好推： dp [ i ] = min { sum[ i ] - sum[ j - 1] } i - j + 1 <= k ,我们可以变一下型，dp [ i ] = sum[ i ] - min { sum[ j - 1 ] } ，这样 sum[ j - 1 ] 就可以用单调队列来维护了。因为是环形的，可以展开成为两倍。

代码：

#include<iostream>#include<sstream>#include<map>#include<cmath>#include<fstream>#include<queue>#include<vector>#include<sstream>#include<cstring>#include<cstdio>#include<stack>#include<bitset>#include<ctime>#include<string>#include<cctype>#include<iomanip>#include<algorithm>using namespace std  ;#define INT long long int#define L(x)  (x * 2)#define R(x)  (x * 2 + 1)const int INF = 0x3f3f3f3f ;const double esp = 0.0000000001 ;const double PI = acos(-1.0) ;const int mod = 1000000007 ;const int MY = (1<<5) + 5 ;const int MX = 200000 + 5 ;const int S = 20 ;int n ,k ,St ,End ,ans ,m ;int g[MX] ,sum[MX] ,deq[MX] ,deqv[MX] ;void input(){   scanf("%d%d" ,&n ,&k) ;   sum[0] = 0 ;   for(int i = 1 ;i <= n ; ++i)   {       scanf("%d" ,&g[i]) ;       g[i+n] = g[i] ;   }   m = n ;   n = n*2 ;   for(int i = 1 ;i <= n ; ++i)      sum[i] = sum[i-1] + g[i] ;}void DP(){    int front = 0 ,end = 0 ,temp ,a1 ,a2 ;    St = INF ; ans = -INF ; End = 0 ;    ans = -INF ;    for(int i = 1 ;i <= n ; ++i)    {        while(front < end && deqv[end-1] > sum[i-1])  // 去除大于等于此元素的值                end-- ;        deq[end] = i-1 ;        deqv[end++] = sum[i-1] ;        temp = sum[i] - deqv[front] ;        if(temp > ans)        {            ans = temp ;            St = deq[front] + 1 ;            End = i ;        }        while(front < end && deq[front] <= i-k)  // 去除冗余元素                  front++ ;    }}int main(){    //freopen("input.txt" ,"r" ,stdin) ;    int Tx ;    scanf("%d" ,&Tx) ;    while(Tx--)    {        input() ;        DP() ;        if(St > m) St -= m ;        if(End > m)  End -= m ;        printf("%d %d %d\n" ,ans ,St ,End) ;    }    return 0 ;}

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