线段树 Codeforces292E Copying Data

时间:2022-11-24 07:19:44


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题意:给两个数组A和B,有两种操作,操作1,将A数组的一部分复制到B数组的某一部分上,操作2,输出B数组的p位置当前的数值。

思路:利用线段树打懒惰标记成段更新裸题

#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck(x) cout<<"["<<x<<"]"
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int>PII;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

const int MX = 1e5 + 5;

int A[MX], B[MX];
struct Data {
int x, y, k;
Data() {}
Data(int _x, int _y, int _k) {
x = _x; y = _y; k = _k;
}
} col[MX << 2];
void push_down(int rt) {
if(col[rt].k) {
col[rt << 1] = col[rt << 1 | 1] = col[rt];
col[rt].k = 0;
}
}
void build(int l, int r, int rt) {
col[rt].k = 0;
if(l == r) return;

int m = (l + r) >> 1;
build(lson);
build(rson);
}
Data query(int x, int l, int r, int rt) {
if(l == r) {
return col[rt];
}
push_down(rt);
int m = (l + r) >> 1;
if(x <= m) return query(x, lson);
else return query(x, rson);
}
void update(int L, int R, Data s, int l, int r, int rt) {
if(L <= l && r <= R) {
col[rt] = s;
return;
}

push_down(rt);
int m = (l + r) >> 1;
if(L <= m) update(L, R, s, lson);
if(R > m) update(L, R, s, rson);
}

int main() {
int n, m; //FIN;
while(~scanf("%d%d", &n, &m)) {
for(int i = 1; i <= n; i++) {
scanf("%d", &A[i]);
}
for(int i = 1; i <= n; i++) {
scanf("%d", &B[i]);
}

build(1, n, 1);
for(int i = 1; i <= m; i++) {
int op, x, y, k;
scanf("%d", &op);
if(op == 1) {
scanf("%d%d%d", &x, &y, &k);
update(y, y + k - 1, Data(x, y, k), 1, n, 1);
} else {
scanf("%d", &k);
Data s = query(k, 1, n, 1);
if(!s.k) printf("%d\n", B[k]);
else printf("%d\n", A[k - s.y + s.x]);
}
}
}
return 0;
}