hdu 3415 单调队列

sum[i]为前缀和

对于以i结尾的一段满足要求的最大和则为x=sum[i]-sum[j-1],i-j+1<=k;

 //#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define pb(a) push_back(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c){return min(min(a,b),min(a,c));}
template<class T> T max(const T& a,const T& b,const T& c){return max(max(a,b),max(a,c));}
void debug()
{
#ifdef ONLINE_JUDGE
#else
    freopen("d:\\in.txt","r",stdin);
    //freopen("d:\\out1.txt","w",stdout);
#endif
}
char getch()
{
    char ch;
    while((ch=getchar())!=EOF)
    {
        if(ch!=' '&&ch!='\n')return ch;
    }
    return EOF;
}
int da[100100],sum[200200],q[200200];
int main()
{
    debug();
    int t;
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++)
    {
        int n,k;
        scanf("%d%d",&n,&k);
        sum[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&da[i]);
            sum[i]=sum[i-1]+da[i];
        }
        for(int i=1;i<=n;i++)
        {
            sum[i+n]=sum[i+n-1]+da[i];
        }
        int front=1,rear=1;
        q[1]=0;
        int maxx=INT_MIN,ml,mr;
        for(int i=1;i<=2*n;i++)
        {
            while(front<=rear&&i-q[front]>k)front++;
            if(sum[i]-sum[q[front]]>maxx)
            {
                maxx=sum[i]-sum[q[front]];
                ml=q[front]+1;mr=i;
            }
            while(front<=rear&&sum[i]<sum[q[rear]])rear--;
            q[++rear]=i;
        }
        if(mr>n)mr=mr%n;
        printf("%d %d %d\n",maxx,ml,mr);
    }
    return 0;
}