Sequence query 好题啊

Problem Description

Given a sequence of N positive numbers，and M queries

A query maybe :

1 x，find the Maximun "weight" of the consecutive subsequence whose length >= x

2 x, find the Minimum length of the consecutive subsequence whose weight >= x

the weight of a consecutive subsequence Seq：weight(Seq) = length of Seq * minimum number in Seq.

Input

The first line is an integer T(1<=T<=100) ，the number of test cases;

For each test case，

the first line contains two integer N，M(1<=N,M<=100000)

the second line contains N positive integers(all between [1,2^31-1])

The following M lines contains queries as descripted above, all number is within signed ilong long

any subsequences should not be empty(length >= 1)

Output

for each query，output a line contains an answer you find，if the answer doesn't exist，output -1

Sample Input

Sample Output

2
1

用单调栈预处理出当以 a[i]为最小值的时候，最左向左延伸到哪里，最右向右延伸到哪里
不妨设为 lt[i],rt[i].
伪代码: a[0] = a[n + 1] = -1;
for(int i = 1; i <= n; i ++) {
lt[i] = i;
while(a[i] <= a[lt[i] - 1]) lt[i] = lt[lt[i] - 1];
}
rt[i]同样处理。当然，有很多种处理方法。
用 max_len[i]表示，当长度为 i 的时候，延伸长度不小于 i 的 ai 的最大值, for(int i = 1; i <= n; i ++) max_len[rt[i] - lt[i] + 1] = max(max_len[rt[i] - lt[i] + 1],a[i]);
max_len[i] = max(max_len[i],max_len[i + 1])
对于第一问，处理一个后缀，dp[i] = max(dp[i + 1],max_len[i] * i)
输入一个 x，判掉不合法的之后，直接输出 dp[x]
对于第二问，和第一问类似，处理一个前缀，然后需要二分。

 1 #include <iostream>
 2 #include <string.h>
 3 #include <stdio.h>
 4 #include <math.h>
 5 #include <stdlib.h>
 6 #include <queue>
 7 using namespace std;
 8 #define ll long long
 9 ll a[110000],lt[110000],rt[110000],n;
10 ll dp[110000],dp1[110000],maxlen[110000];
11 void fun(ll x)
12 {
13     int l,r,ans,m;
14     l=1,r=n,ans=-1;
15     while(l<=r)
16     {
17         m=(l+r)>>1;
18         if(x>dp1[m])
19         {
20             l=m+1;
21         }
22         else
23         {
24             r=m-1;
25             ans=m;
26         }
27     }
28     printf("%d
",ans);
29 }
30 int main()
31 {
32     //freopen("in.txt","r",stdin);
33     int t,i,m;
34     ll x,y;
35     scanf("%d",&t);
36     while(t--)
37     {
38         scanf("%d%d",&n,&m);
39         a[n+1]=a[0]=-1;
40         for(i=1; i<=n; i++)
41             scanf("%lld",&a[i]);
42         for(i=1; i<=n; i++)
43         {
44             lt[i]=i;
45             while(a[i]<=a[lt[i]-1])lt[i]=lt[lt[i]-1];
46         }
47         for(i=n; i>=1; i--)
48         {
49             rt[i]=i;
50             while(a[i]<=a[rt[i]+1])rt[i]=rt[rt[i]+1];
51         }
52         memset(maxlen,0,sizeof(maxlen));
53         for(i=1; i<=n; i++)
54             maxlen[rt[i]-lt[i]+1]=max(maxlen[rt[i]-lt[i]+1],a[i]);
55         for(i=n-1; i>=1; i--)
56             maxlen[i]=max(maxlen[i],maxlen[i+1]);
57         memset(dp,0,sizeof(dp));
58         memset(dp1,0,sizeof(dp1));
59         for(i=n; i>=1; i--)
60             dp[i]=max(dp[i+1],maxlen[i]*i);
61         for(i=1; i<=n; i++)
62             dp1[i]=max(dp1[i-1],maxlen[i]*i);
63 
64         for(i=0; i<m; i++)
65         {
66             scanf("%lld%lld",&x,&y);
67             if(x==1)
68             {
69                 if(y>n)
70                 printf("-1
");
71                 else
72                 {
73                     if(y<=0)
74                     y=1;
75                     printf("%lld
",dp[y]);
76                 }
77                 
78             }
79             else
80             {
81                 fun(y);
82             }
83         }
84     }
85 }

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