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

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 #include 
 6  
 7 using namespace std;
 8  
 9 #define gc getchar
10 #define add(x,y) x=x*10+y-'0'
11  
12 template <class T>
13 __inline void read(T &x)
14 {
15     x = 0; char c = gc();
16      
17     while (c < '0')
18         c = gc();
19      
20     while (c >= '0')add(x,c),
21         c = gc();
22 }
23  
24 #define mem(x) memset(x,0,sizeof(x))
25 #define rep(x) for(int i=1;i<=x;++i)
26  
27 #define ll long long
28 #define LL long long
29  
30 #define N 4000005
31  
32 int n;
33 int d;
34 int num[N];
35 int que[N];
36  
37 LL p;
38 LL sum[N];
39 LL mex[N];
40  
41 signed main(void)
42 {
43     read(n);
44     read(p);
45     read(d);
46      
47     rep(n)read(num[i]);
48     rep(n)sum[i] = sum[i - 1] + num[i];
49     rep(n - d + 1)mex[i] = sum[i + d - 1] - sum[i - 1];
50      
51     int h = 0, t = 0, lt = 0, ans = 0;
52      
53     for (int i = d; i <= n; ++i)
54     {
55         while (h < t && mex[que[t - 1]] <= mex[i - d + 1])
56             --t;
57              
58         que[t++] = i - d + 1;
59          
60         while (sum[i] - sum[lt] - mex[que[h]] > p)
61         {
62             ++lt; while (que[h] <= lt)++h;
63         }
64          
65         if (i - lt > ans)ans = i - lt;
66     }
67      
68     printf("%d\n", ans);
69 }

BZOJ_4385.cpp

 

 1 #include 
 2 
 3 typedef long long longint;
 4 
 5 const int maxn = 4000005;
 6 
 7 int n;
 8 int d;
 9 int num[maxn];
10 int que[maxn];
11 
12 longint p;
13 longint sum[maxn];
14 longint mex[maxn];
15 
16 signed main(void)
17 {
18     scanf("%d%lld%d", &n, &p, &d);
19 
20     for (int i = 1; i <= n; ++i)
21         scanf("%d", num + i);
22 
23     for (int i = 1; i <= n; ++i)
24         sum[i] = sum[i - 1] + num[i];
25 
26     for (int i = 1; i <= n - d + 1; ++i)
27         mex[i] = sum[i + d - 1] - sum[i - 1];
28 
29     int head = 0, tail = 0, left = 0, answer = 0;
30 
31     for (int i = d; i <= n; ++i) {
32         while (head < tail && mex[que[tail - 1]] <= mex[i - d + 1])
33             --tail; // queue pop back
34 
35         que[tail++] = i - d + 1;
36 
37         while (sum[i] - sum[left] - mex[que[head]] > p) {
38             ++left;
39             while (que[head] <= left)
40                 ++head; // queue pop front
41         }
42 
43         if (answer < i - left)
44             answer = i - left;
45     }
46 
47     printf("%d\n", answer);
48 }

BZOJ_4385.cpp

 

@Author: YouSiki

转载于:https://www.cnblogs.com/yousiki/p/6091604.html