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 = ; char c = gc();

     while (c < '')
         c = gc();

     while (c >= '')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 - ] + num[i];
     rep(n - d + )mex[i] = sum[i + d - ] - sum[i - ];

     int h = , t = , lt = , ans = ;

     for (int i = d; i <= n; ++i)
     {
         while (h < t && mex[que[t - ]] <= mex[i - d + ])
             --t;

         que[t++] = i - d + ;

         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);
 }

BZOJ_4385.cpp

 #include <bits/stdc++.h>

 typedef long long longint;

 const int maxn = ;

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

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

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

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

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

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

         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);
 }

BZOJ_4385.cpp

@Author: YouSiki