BZOJ 4385: [POI2015]Wilcze doły

4385: [POI2015]Wilcze doły

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 648  Solved: 263
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Description

给定一个长度为n的序列，你有一次机会选中一段连续的长度不超过d的区间，将里面所有数字全部修改为0。
请找到最长的一段连续区间，使得该区间内所有数字之和不超过p。

Input

第一行包含三个整数n,p,d(1<=d<=n<=2000000，0<=p<=10^16)。
第二行包含n个正整数，依次表示序列中每个数w[i](1<=w[i]<=10^9)。

Output

包含一行一个正整数，即修改后能找到的最长的符合条件的区间的长度。

Sample Input

9 7 2
3 4 1 9 4 1 7 1 3

Sample Output

5

HINT

 

将第4个和第5个数修改为0，然后可以选出区间[2,6]，总和为4+1+0+0+1=6。

 

Source

鸣谢Claris

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分析

显然单调队列O(N)扫过去即可，就是注意优化常数。

代码

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
 
using namespace std;
 
#define gc getchar
#define add(x,y) x=x*10+y-'0'
 
template <class T>
__inline void read(T &x)
{
    x = 0; char c = gc();
     
    while (c < '0')
        c = gc();
     
    while (c >= '0')add(x,c),
        c = gc();
}
 
#define mem(x) memset(x,0,sizeof(x))
#define rep(x) for(int i=1;i<=x;++i)
 
#define ll long long
#define LL long long
 
#define N 4000005
 
int n;
int d;
int num[N];
int que[N];
 
LL p;
LL sum[N];
LL mex[N];
 
signed main(void)
{
    read(n);
    read(p);
    read(d);
     
    rep(n)read(num[i]);
    rep(n)sum[i] = sum[i - 1] + num[i];
    rep(n - d + 1)mex[i] = sum[i + d - 1] - sum[i - 1];
     
    int h = 0, t = 0, lt = 0, ans = 0;
     
    for (int i = d; i <= n; ++i)
    {
        while (h < t && mex[que[t - 1]] <= mex[i - d + 1])
            --t;
             
        que[t++] = i - d + 1;
         
        while (sum[i] - sum[lt] - mex[que[h]] > p)
        {
            ++lt; while (que[h] <= lt)++h;
        }
         
        if (i - lt > ans)ans = i - lt;
    }
     
    printf("%d\n", ans);
}

 

#include <bits/stdc++.h>

typedef long long longint;

const int maxn = 4000005;

int n;
int d;
int num[maxn];
int que[maxn];

longint p;
longint sum[maxn];
longint mex[maxn];

signed main(void)
{
    scanf("%d%lld%d", &n, &p, &d);

    for (int i = 1; i <= n; ++i)
        scanf("%d", num + i);

    for (int i = 1; i <= n; ++i)
        sum[i] = sum[i - 1] + num[i];

    for (int i = 1; i <= n - d + 1; ++i)
        mex[i] = sum[i + d - 1] - sum[i - 1];

    int head = 0, tail = 0, left = 0, answer = 0;

    for (int i = d; i <= n; ++i) {
        while (head < tail && mex[que[tail - 1]] <= mex[i - d + 1])
            --tail; // queue pop back

        que[tail++] = i - d + 1;

        while (sum[i] - sum[left] - mex[que[head]] > p) {
            ++left;
            while (que[head] <= left)
                ++head; // queue pop front
        }

        if (answer < i - left)
            answer = i - left;
    }

    printf("%d\n", answer);
}

 

@Author: YouSiki