codeforces 232D Fence
John Doe has a crooked fence, consisting of n rectangular planks, lined up from the left to the right: the plank that goes i-th (1 ≤ i ≤ n)(from left to right) has width 1 and height hi. We will assume that the plank that goes i-th (1 ≤ i ≤ n) (from left to right) has index i.
A piece of the fence from l to r (1 ≤ l ≤ r ≤ n) is a sequence of planks of wood with indices from l to r inclusive, that is, planks with indices l, l + 1, ..., r. The width of the piece of the fence from l to r is value r - l + 1.
Two pieces of the fence from l1 to r1 and from l2 to r2 are called matching, if the following conditions hold:
- the pieces do not intersect, that is, there isn't a single plank, such that it occurs in both pieces of the fence;
- the pieces are of the same width;
- for all i (0 ≤ i ≤ r1 - l1) the following condition holds: hl1 + i + hl2 + i = hl1 + hl2.
John chose a few pieces of the fence and now wants to know how many distinct matching pieces are for each of them. Two pieces of the fence are distinct if there is a plank, which belongs to one of them and does not belong to the other one.
The first line contains integer n (1 ≤ n ≤ 105) — the number of wood planks in the fence. The second line contains n space-separated integers h1, h2, ..., hn (1 ≤ hi ≤ 109) — the heights of fence planks.
The third line contains integer q (1 ≤ q ≤ 105) — the number of queries. Next q lines contain two space-separated integers li and ri(1 ≤ li ≤ ri ≤ n) — the boundaries of the i-th piece of the fence.
For each query on a single line print a single integer — the number of pieces of the fence that match the given one. Print the answers to the queries in the order, in which the queries are given in the input.
10
1 2 2 1 100 99 99 100 100 100
6
1 4
1 2
3 4
1 5
9 10
10 10
1
2
2
0
2
9 题解:http://tieba.baidu.com/p/2114943791?pid=28521015525&see_lz=1
code:
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#define maxn 200005
using namespace std;
char ch;
int n,q,N,m,l,r,h[maxn],d[maxn],s[maxn],st[][maxn];
int SA[maxn],rank[maxn],t1[maxn],t2[maxn],height[maxn],sum[maxn];
struct DATA{
int v,id;
}list[maxn];
bool cmp(DATA a,DATA b){
if (a.v!=b.v) return a.v<b.v;
return a.id<b.id;
}
struct seg{
int tot,son[maxn*][],cnt[maxn*];
void init(){tot=N;}
void insert(int k,int p,int l,int r,int x){
if (l==r){cnt[k]=cnt[p]+;return;}
int m=(l+r)>>;
if (x<=m){
cnt[k]=cnt[p]+,son[k][]=++tot,son[k][]=son[p][];
insert(son[k][],son[p][],l,m,x);
}
else{
cnt[k]=cnt[p]+,son[k][]=son[p][],son[k][]=++tot;
insert(son[k][],son[p][],m+,r,x);
}
}
int query(int k,int l,int r,int x,int y){
if (!k||x>y) return ;
if (l==x&&r==y) return cnt[k];
int m=(l+r)>>;
if (y<=m) return query(son[k][],l,m,x,y);
else if (x<=m) return query(son[k][],l,m,x,m)+query(son[k][],m+,r,m+,y);
else return query(son[k][],m+,r,x,y);
}
int query(int l,int r,int x,int y){return query(r,,N,x,y)-query(l-,,N,x,y);}
}T;
bool ok;
void read(int &x){
for (ok=,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=;
for (x=;isdigit(ch);x=x*+ch-'',ch=getchar());
if (ok) x=-x;
}
void get_SA(){
for (int i=;i<=N;i++) list[i]=(DATA){s[i],i};
sort(list+,list+N+,cmp);
for (int i=;i<=N;i++) SA[i]=list[i].id;
int *x=t1,*y=t2,tot=;
x[SA[]]=m=;
for (int i=;i<=N;i++){
if (s[SA[i]]!=s[SA[i-]]) m++;
x[SA[i]]=m;
}
for (int len=;tot<N;len<<=,m=tot){
tot=;
for (int i=N-len+;i<=N;i++) y[++tot]=i;
for (int i=;i<=N;i++) if (SA[i]>len) y[++tot]=SA[i]-len;
for (int i=;i<=m;i++) sum[i]=;
for (int i=;i<=N;i++) sum[x[y[i]]]++;
for (int i=;i<=m;i++) sum[i]+=sum[i-];
for (int i=N;i>=;i--) SA[sum[x[y[i]]]--]=y[i];
swap(x,y),x[SA[]]=tot=;
for (int i=;i<=N;i++){
if (y[SA[i]]!=y[SA[i-]]||y[SA[i]+len]!=y[SA[i-]+len]) tot++;
x[SA[i]]=tot;
}
}
for (int i=;i<=N;i++) rank[i]=x[i];
}
void get_height(){
for (int i=,j=;i<=N;i++){
if (rank[i]==) continue;
while (s[i+j]==s[SA[rank[i]-]+j]) j++;
height[rank[i]]=j;
if (j>) j--;
}
}
void prepare(){
for (int i=;i<=N;i++) st[][i]=height[i];
for (int i=;i<=;i++)
for (int j=;j<=N;j++){
st[i][j]=st[i-][j];
if (j+(<<(i-))<=N) st[i][j]=min(st[i][j],st[i-][j+(<<(i-))]);
}
T.init();
for (int i=;i<=N;i++) T.insert(i,i-,,N,SA[i]);
}
int calc(int l,int r){
if (l>r) swap(l,r);
int t=; l++;
if (l==r) return height[r];
for (;l+(<<t)<r;t++);
if (l+(<<t)>r) t--;
return min(st[t][l],st[t][r-(<<t)+]);
}
int find(int s,int x,int op){
int l,r,m;
if (op) l=s,r=N;else l=,r=s;
while (l!=r){
m=(l+r)>>;
if (op) m++;
if (calc(m,s)<x){
if (op) r=m-;
else l=m+;
}
else{
if (op) l=m;
else r=m;
}
}
return l;
}
void query(int l,int r){
int x=find(rank[l],r-l,),y=find(rank[l],r-l,);
printf("%d\n",T.query(x,y,n+,n+(l-)-(r-l))+T.query(x,y,n+(r+),N));
}
int main(){
read(n);
for (int i=;i<=n;i++) read(h[i]);
for (int i=;i<n;i++) d[i]=h[i+]-h[i];
for (int i=;i<n;i++) s[i]=d[i];
s[n]=;
for (int i=;i<n;i++) s[n+i]=-d[i];
N=(n<<)-;
get_SA(),get_height(),prepare();
for (read(q);q;q--){
read(l),read(r);
if (l==r) printf("%d\n",n-);
else query(l,r);
}
return ;
}
codeforces 232D Fence的更多相关文章
- codeforces 448CPainting Fence
版权声明:本文为博主原创文章,未经博主同意不得转载. https://blog.csdn.net/y990041769/article/details/37935237 题目:codeforces 4 ...
- Fence(codeforces 232D)
题意: 对于给定的a[1..n],定义区间[s,t]和[x,y]"匹配"当且仅当下列条件同时满足:1. t-s=y-x,即长度相同.3. t<x或s>y,即两区间没有交 ...
- CodeForces 363B Fence
Fence Time Limit: 1000ms Memory Limit: 262144KB This problem will be judged on CodeForces. Original ...
- codeforces B.Fence 解题报告
题目链接:http://codeforces.com/problemset/problem/363/B 题目意思:给定整数n和k,需要从n个数中找出连续的k个数之和最小,输出这连续的k个数中的第一个数 ...
- Codeforces 659G Fence Divercity dp
Fence Divercity 我们设a[ i ] 为第 i 个围栏被切的最靠下的位置, 我们发现a[ i ] 的最大取值有一下信息: 如果从i - 1过来并在 i 结束a[ i ] = min(h ...
- codeforces 349B Color the Fence 贪心,思维
1.codeforces 349B Color the Fence 2.链接:http://codeforces.com/problemset/problem/349/B 3.总结: 刷栅栏.1 ...
- Codeforces 484E Sign on Fence(是持久的段树+二分法)
题目链接:Codeforces 484E Sign on Fence 题目大意:给定给一个序列,每一个位置有一个值,表示高度,如今有若干查询,每次查询l,r,w,表示在区间l,r中, 连续最长长度大于 ...
- CF&&CC百套计划4 Codeforces Round #276 (Div. 1) E. Sign on Fence
http://codeforces.com/contest/484/problem/E 题意: 给出n个数,查询最大的在区间[l,r]内,长为w的子区间的最小值 第i棵线段树表示>=i的数 维护 ...
- Codeforces Round #355 (Div. 2) A. Vanya and Fence 水题
A. Vanya and Fence 题目连接: http://www.codeforces.com/contest/677/problem/A Description Vanya and his f ...
随机推荐
- 【行业干货】ASOS:外来快时尚品牌的入华战 - 行业干货 - 京东内部论坛 - Powered by Discuz!
[行业干货]ASOS:外来快时尚品牌的入华战 - 行业干货 - 京东内部论坛 - Powered by Discuz! [行业干货]ASOS:外来快时尚品牌的入华战
- 【纯干货】SVN使用时应注意的那些事
一.SVN使用步骤 检出 checkout 更新 update 冲突 confilicte 添加 Add (没有添加项目可不写) 填写svn日志 提交 commit你以为到这儿就结束了吗?....NO ...
- JavaScript使用需要注意的细节
1.JavaScript区分大小写 在JavaScript中对象,变量.函数都是区分大小写的,例如: Object表示对象,Arrary表示数组,而写成object,arrary的时候JavaScri ...
- [置顶] C语言单元测试框架
unitest.h /****************************************************************************** * * * This ...
- Oracle MERGE INTO的使用方法
非常多时候我们会出现例如以下情境,假设一条数据在表中已经存在,对其做update,假设不存在,将新的数据插入.假设不使用Oracle提供的merge语法的话,可能先要上数据库select查询一下看是否 ...
- C++类中静态变量
以下是对类中static变量的一点解说 =============================================== 静态数据成员的用法和注意事项例如以下: ...
- Ext入门的第一个程序(1)
1.Ext是什么? extjs是集UI和ajax框架与一身的,界面又好看,又有很强的ajax交互功能,适合不会做漂亮页面的程序员用的,缺点就是太大了,要导入近800KB左右的js和css文件,这对于w ...
- 从 ReactiveCocoa 中能学到什么?不用此库也能学以致用
从知道ReactiveCocoa开始就发现对这个库有不同的声音,上次参加<T>技术沙龙时唐巧对在项目中已全面使用FRP的代码家提出为什么这种编程模型出现了这么长时间怎么像ReactiveC ...
- Android ViewPager 打造炫酷欢迎页
Android ViewPager 打造炫酷欢迎页 ViewPager是Android扩展v4包中的类,这个类可以让用户切换当前的View.对于这个类的应用场景,稍加修改就可以应用到多个环境下.比如: ...
- iOS报错Expected selector for Objective-C method
这个报错非常恶心:原因竟然是在导入头文件的地方多写了一个"+"号,可能问题在一个文件,报错在另一个文件