Croc Champ 2013 - Round 1 E. Copying Data 线段树
2016-09-15 11:59
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题目链接:
http://codeforces.com/problemset/problem/292/EE. Copying Data
time limit per test2 secondsmemory limit per test256 megabytes问题描述
We often have to copy large volumes of information. Such operation can take up many computer resources. Therefore, in this problem you are advised to come up with a way to copy some part of a number array into another one, quickly.
More formally, you've got two arrays of integers a1, a2, ..., an and b1, b2, ..., bn of length n. Also, you've got m queries of two types:
Copy the subsegment of array a of length k, starting from position x, into array b, starting from position y, that is, execute by + q = ax + q for all integer q (0 ≤ q < k). The given operation is correct — both subsegments do not touch unexistent elements.
Determine the value in position x of array b, that is, find value bx.
For each query of the second type print the result — the value of the corresponding element of array b.
输入
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of elements in the arrays and the number of queries, correspondingly. The second line contains an array of integers a1, a2, ..., an (|ai| ≤ 109). The third line contains an array of integers b1, b2, ..., bn (|bi| ≤ 109).
Next m lines contain the descriptions of the queries. The i-th line first contains integer ti — the type of the i-th query (1 ≤ ti ≤ 2). If ti = 1, then the i-th query means the copying operation. If ti = 2, then the i-th query means taking the value in array b. If ti = 1, then the query type is followed by three integers xi, yi, ki (1 ≤ xi, yi, ki ≤ n) — the parameters of the copying query. If ti = 2, then the query type is followed by integer xi (1 ≤ xi ≤ n) — the position in array b.
All numbers in the lines are separated with single spaces. It is guaranteed that all the queries are correct, that is, the copying borders fit into the borders of arrays a and b.
输出
For each second type query print the result on a single line.
样例输入
5 10
1 2 0 -1 3
3 1 5 -2 0
2 5
1 3 3 3
2 5
2 4
2 1
1 2 1 4
2 1
2 4
1 4 2 1
2 2
样例输出
0
3
-1
3
2
3
-1
题意
给你大小为n的两个数组,a,b;
执行两个操作:
1,x,y,k,把a数组中以x开头长度为k的串覆盖到b数组中以y开头的长度为k的位置。
2,x, 查询b[x];
题解
对于查询1在b数组上做成段覆盖更新:(y,y+k-1),值为查询的id。然后对于查询2做单点查询,查到对应的查询再去a数组中找对应的位置。
代码
#include<map> #include<set> #include<cmath> #include<queue> #include<stack> #include<ctime> #include<vector> #include<cstdio> #include<string> #include<bitset> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #include<functional> using namespace std; #define X first #define Y second #define mkp make_pair #define lson (o<<1) #define rson ((o<<1)|1) #define mid (l+(r-l)/2) #define sz() size() #define pb(v) push_back(v) #define all(o) (o).begin(),(o).end() #define clr(a,v) memset(a,v,sizeof(a)) #define bug(a) cout<<#a<<" = "<<a<<endl #define rep(i,a,b) for(int i=a;i<(b);i++) #define scf scanf #define prf printf typedef long long LL; typedef vector<int> VI; typedef pair<int,int> PII; typedef vector<pair<int,int> > VPII; const int INF=0x3f3f3f3f; const LL INFL=0x3f3f3f3f3f3f3f3fLL; const double eps=1e-8; const double PI = acos(-1.0); //start---------------------------------------------------------------------- const int maxn=1e5+10; int setv[maxn<<2]; void pushdown(int o){ if(setv[o]!=-1){ setv[lson]=setv[rson]=setv[o]; setv[o]=-1; } } int qp,qv; void query(int o,int l,int r){ if(l==r||setv[o]!=-1){ qv=setv[o]; }else{ if(qp<=mid) query(lson,l,mid); else query(rson,mid+1,r); } } int ul,ur,uv; void update(int o,int l,int r){ if(ul<=l&&r<=ur){ setv[o]=uv; }else{ pushdown(o); if(ul<=mid) update(lson,l,mid); if(ur>mid) update(rson,mid+1,r); } } struct Node{ int x,y,k; Node(int x,int y,int k):x(x),y(y),k(k){} Node(){} }que[maxn]; int a[maxn],b[maxn]; int n,m; void init(){ clr(setv,-1); } int main() { scf("%d%d",&n,&m); init(); for(int i=1;i<=n;i++) scf("%d",&a[i]); for(int i=1;i<=n;i++) scf("%d",&b[i]); ul=1,ur=n,uv=0; update(1,1,n); for(int i=1;i<=m;i++){ int cmd; scf("%d",&cmd); if(cmd==1){ int x,y,k; scf("%d%d%d",&x,&y,&k); que[i]=Node(x,y,k); ul=y,ur=y+k-1,uv=i; update(1,1,n); }else{ int x; scf("%d",&x); qp=x; query(1,1,n); int ans; if(qv==0) ans=b[x]; else ans=a[x-que[qv].y+que[qv].x]; prf("%d\n",ans); } } return 0; } //end-----------------------------------------------------------------------
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