E. GukiZ and GukiZianap平方分桶

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：给n个数，q个询问，包括两种操作。第一种是l，r的区间所有数加上x。第二种操作是查找整个区间里面等于x相距最远的两个位置的差值。即xi=xj=y,求最大的j-i。

解法：因为涉及区间操作起先一开始想到的就是线段树，只是因为需要查找y的位置，线段树似乎并不能办到。于是接着转而想到了平方分桶，即是将整个区间分成sqrt(n)个桶，每个桶之中存储sqrt(n)个数。将桶内的元素排序，给每一个桶一个add值。表示整个桶的区间增加的数，如果对整个桶操作，直接对这个标记进行操作即可。如果是只有桶内一部分元素进行操作，直接遍历，因为桶内元素最多只有sqrt(n)个，所以大大降低了复杂度。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <cmath>
#include <stack>
#include <cstdlib>
#include <string>
#include <map>
#include <set>
#include <algorithm>
#include <functional>
#define rep(i,a,b) for (int i=a;i<((b)+1);i++)
#define Rep(i,a,b) for (int i=a;i>=b;i--)
#define foreach(e,x) for (__typeof(x.begin()) e=x.begin();e!=x.end();e++)
#define mid ((l+r)>>1) //segment tree
#define lson (k<<1) //segment tree
#define rson (k<<1|1) //segment tree
#define MEM(a,x) memset(a,x,sizeof a)
#define L ch[r][0] //splay tree
#define R ch[r][1] //splay tree
#define keyvalue ch[ch[root][1]][0] //splay tree
#define eps 1e-6
using namespace std;
const int N=600050;
const long long Mod=1e9+7;
const int inf=0x3f3f3f3f;
typedef pair<int, int> pii;
typedef long long ll;
typedef pair<ll,ll> pll;
typedef pair<ll,int> pli;
ll a
,add
;
vector<ll> vt
;
int n,q,bk=1000,pos
,bn=1;
bool cmp(const int &x,const int &y) {
return a[x]==a[y]?x<y:a[x]<a[y];
}
void update(int l,int r,ll x) {
int _l=pos[l],_r=pos[r];
if (_l==_r) {
rep(i,l,r) a[i]+=x;
sort(vt[_l].begin(),vt[_l].end(),cmp);
return ;
}
if (pos[l]==pos[l-1]) _l++;
if (pos[r]==pos[r+1]) _r--;
rep(i,_l,_r) add[i]+=x;
if (pos[l]==pos[l-1]) {
for (int i=l;pos[i]==_l-1;i++) a[i]+=x;
sort(vt[_l-1].begin(),vt[_l-1].end(),cmp);
}
if (pos[r]==pos[r+1]) {
for (int i=r;pos[i]==_r+1;i--) a[i]+=x;
sort(vt[_r+1].begin(),vt[_r+1].end(),cmp);
}
}
int query(ll x) {
int l=-1,r=-1;
rep(i,1,bn) {
a[0]=x-add[i];
vector<ll>::iterator it=lower_bound(vt[i].begin(),vt[i].end(),0,cmp);
if (it==vt[i].end()) continue;
if (a[*it]+add[i]==x) {
l=*it;
break;
}
}
if (l==-1) return -1;
Rep(i,bn,1) {
a[n+1]=x-add[i];
vector<ll>::iterator it=lower_bound(vt[i].begin(),vt[i].end(),n+1,cmp);
if (it==vt[i].begin()) continue;
it--;
if (a[*it]+add[i]==x) {
r=*it;
break;
}
}
return r-l;
}
int main() {
scanf("%d%d",&n,&q);
bn=(n/1000)+(n%1000>0);
rep(i,1,n) {
scanf("%lld",&a[i]);
pos[i]=(i-1)/1000+1;
vt[pos[i]].push_back(i);
}
rep(i,1,bn) sort(vt[i].begin(),vt[i].end(),cmp);
rep(i,1,q) {
int sw,l,r;
ll x;
scanf("%d",&sw);
if (sw==1) {
scanf("%d%d%lld",&l,&r,&x);
update(l,r,x);
}else {
scanf("%lld",&x);
printf("%d\n",query(x));
}
}
return 0;
}