Codeforces 292 E. Copying Data （线段树）

Description

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 a 1 ,   a 2 ,   . . . ,   a n and b1, b2, ..., bn b 1 ,   b 2 ,   . . . ,   b n of length n n . Also, you’ve got mmCopy the subsegment of array a a of length kk, starting from position x x , into array bb, starting from position y y , that is, execute by + q = ax + qby + q = ax + q for all integer q q (0 ≤ q < k)(0 ≤ q < k). The given operation is correct — both subsegments do not touch unexistent elements.Determine the value in position x x of array bb, that is, find value bx b x For each query of the second type print the result — the value of the corresponding element of array b b .

Input

The first line contains two space-separated integers nn and m m (1 ≤ n, m ≤ 105)(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 a 1 ,   a 2 ,   . . . ,   a n (|ai| ≤ 109) ( | a i |   ≤   10 9 ) . The third line contains an array of integers b1, b2, ..., bn b 1 ,   b 2 ,   . . . ,   b n (|bi| ≤ 109) ( | b i |   ≤   10 9 ) .Next m m lines contain the descriptions of the queries. The i-th line first contains integer titi — the type of the i-th query (1 ≤ ti ≤ 2) ( 1   ≤   t i   ≤   2 ) . If ti = 1 t i   =   1 , then the i-th query means the copying operation. If ti = 2 t i   =   2 , then the i-th query means taking the value in array b b . If ti = 1ti = 1, then the query type is followed by three integers xi, yi, ki x i ,   y i ,   k i (1 ≤ xi, yi, ki ≤ n) ( 1   ≤   x i ,   y i ,   k i   ≤   n ) — the parameters of the copying query. If ti = 2 t i   =   2 , then the query type is followed by integer xi x i (1 ≤ xi ≤ n) ( 1   ≤   x i   ≤   n ) — the position in array b 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 aa and b b .

Output

For each second type query print the result on a single line.

Examples input

5 101 2 0 -1 33 1 5 -2 02 51 3 3 32 52 42 11 2 1 42 12 41 4 2 12 2

Examples output

03-1323-1

题意

对于数组 aa 和数组 b b

用 aa 中从 x x 开始的连续 kk 个元素覆盖掉 b b 中从 yy 开始的连续 k k询问 bb 中下标为 k k $k$ 的元素是多少

思路

我们知道，覆盖算是一种擦写不可逆式的操作，即当前元素的值只与该位置最后一次的覆盖有关

于是结合线段树区间 set 操作，我们在覆盖时将某个区间更新为该次操作的索引

对于每一次的查询，只需要找到其覆盖的索引，然后便可以用公式得出它的值是多少啦~

算是一道线段树的模板题吧~ 当然也有其他做法啦~

AC 代码

#include#define IO ios::sync_with_stdio(false);\ cin.tie(0);\ cout.tie(0);#define default_setv -1using namespace std;typedef long long LL;const int maxn = 2e5+10;int sumv[maxn<<2]; //四倍空间int minv[maxn<<2];int maxv[maxn<<2];int addv[maxn<<2];int setv[maxn<<2]; //初始值：-1int a[maxn];int A[maxn],B[maxn];void maintain(int o,int L,int R){ int lc = o<<1,rc=o<<1|1; sumv[o] = minv[o] = maxv[o] = 0; if(setv[o] != default_setv) { sumv[o] = setv[o] * (R-L+1); minv[o] = maxv[o] = setv[o]; } else if(R>L) { sumv[o] = sumv[lc] + sumv[rc]; minv[o] = min(minv[lc],minv[rc]); maxv[o] = max(maxv[lc],maxv[rc]); } minv[o] += addv[o]; maxv[o] += addv[o]; sumv[o] += addv[o] * (R-L+1);}void pushdown(int o){ int lc = o<<1,rc = o<<1|1; if(setv[o] != default_setv) { setv[lc] = setv[rc] = setv[o]; addv[lc] = addv[rc] = 0; setv[o] = default_setv; } if(addv[o] != 0) { addv[lc] += addv[o]; addv[rc] += addv[o]; addv[o] = 0; }}/* * o: 当前节点(1)，修改区间：[y1,y2] ，添加值：v_add，总区间：[L,R] (这里的L,R必须是2的幂次) */void update_add(int o,int L,int R,int y1,int y2,int v_add){ int lc = o<<1,rc = o<<1|1; if(y1<=L&&y2>=R) addv[o] += v_add; else { pushdown(o); int M = L + (R-L)/2; if(y1<=M) update_add(lc,L,M,y1,y2,v_add); else maintain(lc,L,M); if(y2>M) update_add(rc,M+1,R,y1,y2,v_add); else maintain(rc,M+1,R); } maintain(o,L,R);}/* * o: 当前节点(1)，修改区间：[y1,y2] ，修改值：v_set，总区间：[L,R] (这里的L,R必须是2的幂次) */void update_set(int o, int L, int R,int y1,int y2,int v_set){ int lc = o<<1, rc = o<<1|1; if(y1 <= L && y2 >= R) { setv[o] = v_set; addv[o] = 0; } else { pushdown(o); int M = L + (R-L)/2; if(y1 <= M) update_set(lc, L, M,y1,y2,v_set); else maintain(lc, L, M); if(y2 > M) update_set(rc, M+1, R,y1,y2,v_set); else maintain(rc, M+1, R); } maintain(o, L, R);}/* * o: 当前节点(1)，查询区间：[y1,y2] ，初始累加(0)，总区间：[L,R] (这里的L,R必须是2的幂次) */int _min, _max, _sum;void query(int o, int L, int R, int y1,int y2, int add){ if(setv[o] != default_setv) { _sum += (add+setv[o]+addv[o]) * (min(R, y2)-max(L, y1)+1); _min = min(_min, setv[o]+addv[o]+add); _max = max(_max, setv[o]+addv[o]+add); } else if(y1 <= L && y2 >= R) { _sum += sumv[o] + add * (R-L+1); _min = min(_min, minv[o]+add); _max = max(_max, maxv[o]+add); } else { int M = L + (R-L)/2; if(y1 <= M) query(o*2, L, M, y1, y2,add+addv[o]); if(y2 > M) query(o*2+1, M+1, R, y1, y2, add + addv[o]); }}/* * o: 当前节点(1)，总区间：[L,R] (这里的L,R必须是2的幂次) * 所有 setv 初始化为 -1，所有叶子节点 addv 为当前值 */void build(int o,int L,int R){ if(L==R) { setv[o] = default_setv; addv[o] = sumv[o] = maxv[o] = minv[o] = a[L]; } else { int M = L + (R-L)/2; int lc = o<<1,rc = o<<1|1; build(lc,L,M); build(rc,M+1,R); addv[o] = 0; setv[o] = default_setv; sumv[o] = sumv[lc] + sumv[rc]; maxv[o] = max(maxv[lc],maxv[rc]); minv[o] = min(minv[lc],minv[rc]); }}/* * 测试函数，R 代表叶子节点个数 */void test(int R,int n){ cout<<"----------"< "<<_sum<<"\t "<<_min<<"\t "<<_max<>n>>m; int R = 1; while(R>A[i]; for(int i=1; i<=n; i++) cin>>B[i]; for(int i=1; i<=m; i++) { int op; cin>>op; if(op==2) { cin>>k[i]; _sum = 0; query(1,1,R,k[i],k[i],0); if(_sum==0) cout<>x[i]>>y[i]>>k[i]; update_set(1,1,R,y[i],y[i]+k[i]-1,i); } } return 0;}

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