Croc Champ 2013 - Round 1 E. Copying Data 线段树

题目链接：

http://codeforces.com/problemset/problem/292/E

E. Copying Data

time limit per test2 seconds
memory 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-----------------------------------------------------------------------