Codeforces 551E GukiZ and GukiZiana (分块)

E. GukiZ and GukiZiana

time limit per test
10 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Professor GukiZ was playing with arrays again and accidentally discovered new function, which he called GukiZiana. For given array a,
indexed with integers from 1 to n,
and number y, GukiZiana(a, y) represents
maximum value of j - i, such that aj = ai = y.
If there is no y as an element in a,
then GukiZiana(a, y) is equal to  - 1.
GukiZ also prepared a problem for you. This time, you have two types of queries:

First type has form 1 l r x and
asks you to increase values of all ai such
that l ≤ i ≤ r by the non-negative integer x.

Second type has form 2 y and
asks you to find value of GukiZiana(a, y).

For each query of type 2, print the answer and make GukiZ happy!

Input

The first line contains two integers n, q (1 ≤ n ≤ 5 * 105, 1 ≤ q ≤ 5 * 104),
size of array a, and the number of queries.

The second line contains n integers a1, a2, ... an (1 ≤ ai ≤ 109),
forming an array a.

Each of next q lines contain either four or two numbers, as described in statement:

If line starts with 1, then the query looks like 1 l r x (1 ≤ l ≤ r ≤ n, 0 ≤ x ≤ 109),
first type query.

If line starts with 2, then th query looks like 2 y (1 ≤ y ≤ 109),
second type query.

Output

For each query of type 2, print the value of GukiZiana(a, y),
for y value for that query.

Sample test(s)

input

4 3
1 2 3 4
1 1 2 1
1 1 1 1
2 3

output

2

input

2 3
1 21 2 2 1
2 3
2 4

output

0
-1

题意：

维护一个长为n的序列，支持两种操作：1.将[l,r]内的每个数增加x，2.对于给定的y，输出满足a[j]=a[i]=y的j-i的最大值，无解输出-1.

分析：

按sqrt(n)的大小分块，对于每一个块，维护块内元素的一个有序序列，对于操作1，如果整个块被修改操作覆盖，由于块内元素相对大小不变，只需打一个标记，如果某个块中有部分元素被修改，则暴力重构整个块，询问时对每个块分别查询，最后统计全局最优值。复杂度O(nlogn+qsqrt(n)logn)。

代码：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
ll a[500005];
vector<pair<ll,int> >s[805];
vector<int>res;
vector<pair<ll,int> >::iterator itr;
ll tag[805];
int main()
{
int n,q;
scanf("%d%d",&n,&q);
int b_s=(int)sqrt(n);
for(int i=0;i<n;i++)
{
scanf("%I64d",&a[i]);
}
int loc=-1;
for(int i=0;i<n;i++)
{
if(i%b_s==0)loc++;
s[loc].push_back(make_pair(a[i],i));
}
for(int i=0;i<=loc;i++)
{
sort(s[i].begin(),s[i].end());
}
int op,l,r;
ll x,y;
while(q--)
{
scanf("%d",&op);
if(op==1)
{
scanf("%d%d%I64d",&l,&r,&x);
l--,r--;
int bl=l/b_s;
int br=r/b_s;
for(int i=bl+1;i<br;i++)
{
tag[i]+=x;
}
for(int i=0;i<s[bl].size();i++)
{
s[bl][i].first+=tag[bl];
if(s[bl][i].second>=l && s[bl][i].second<=r)
s[bl][i].first+=x;
}
sort(s[bl].begin(),s[bl].end());
tag[bl]=0;
if(br>bl)
{
for(int i=0;i<s[br].size();i++)
{
s[br][i].first+=tag[br];
if(s[br][i].second>=l && s[br][i].second<=r)
s[br][i].first+=x;
}
sort(s[br].begin(),s[br].end());
tag[br]=0;
}
}
else
{
scanf("%I64d",&y);
res.clear();
for(int i=0;i<=loc;i++)
{
itr=lower_bound(s[i].begin(),s[i].end(),make_pair(y-tag[i],0));
if(itr!=s[i].end() && (*itr).first==y-tag[i])res.push_back((*itr).second);
itr=lower_bound(s[i].begin(),s[i].end(),make_pair(y-tag[i]+1,0));
if(itr!=s[i].begin())
{
itr--;
if((*itr).first==y-tag[i])res.push_back((*itr).second);
}
}
if(res.empty())printf("-1\n");
else
{
sort(res.begin(),res.end());
printf("%d\n",res[res.size()-1]-res[0]);
}
}
}
return 0;
}