cogs 920. [東方S1] 琪露诺

二次联通门 : cogs 920. [東方S1] 琪露诺

/*
    cogs 920. [東方S1] 琪露诺

    dp
    
    方程为dp[i] = max (dp[i - L], dp[i - L + 1] .... dp[i - R - 1], dp[i - R]);
    
    要用单调队列优化
    
    记录路径是只需像最短路问题一样记录前驱
    后递归打印即可 
*/
#include <cstring>
#include <cstdio>

void read (int &now)
{
    register char word = getchar ();
    bool temp = false;
    for (now = 0; word < '0' || word > '9'; word = getchar ())
        if (word == '-')
            temp = true;
    for (; word >= '0' && word <= '9'; now = now * 10 + word - '0', word = getchar ());
    if (temp)
        now = -now;
}

//#define Online
#define INF 1e9

int N, L, R, Answer = -INF;

int pos;
#define Max 1000001
int dp[Max];

int Queue[Max];
int Head, Tail;

int value[Max], pre[Max];

void Print_road (int now)
{
    if (pre[now] != -1)
        Print_road (pre[now]);
    printf ("%d ", now);
}


int main (int argc, char *argv[])
{
    
#ifdef Online

    freopen ("iceroad.in", "r", stdin);
    freopen ("iceroad.out", "w", stdout);

#endif

    read (N);
    read (L);
    read (R);
    
    for (int i = 0; i <= N; i ++)
        read (value[i]);
    Tail = 1;
        
    memset (pre, -1, sizeof pre);
    for (register int i = L, res; i <= N; i ++)
    {
        if (i - L <= 0)
            continue;
        res = dp[i - L];
        for (; Head <= Tail && (Queue[Head] + L > i || (Queue[Head] + R < i) || (Queue[Head] > i)); Head ++);
        for (; Head <= Tail && (res > dp[Queue[Tail]]); Tail --);
        Queue[++ Tail] = i - L;
        pre[i] = Queue[Head];
        dp[i] = value[i] + dp[Queue[Head]];
        if (Answer < dp[i])
        {
            Answer = dp[i];
            pos = i;
        }
    }
    
    printf ("%d
", Answer);
    Print_road (pos);
    printf ("-1");
     
    return 0;
}