hdu--3468(线段树+lazy思想)

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 82238 Accepted: 25451Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

从网上找到了人家对lazy思想的解释,非常容易理解:比如现在需要对[a,b]区间值进行加c操作，那么就从根节点[1,n]开始调用update函数进行操作，如果刚好执行到一个子节点，它的节点标记为rt，这时tree[rt].l == a && tree[rt].r == b 这时我们可以一步更新此时rt节点的sum[rt]的值，sum[rt] += c * (tree[rt].r - tree[rt].l + 1)，注意关键的时刻来了，如果此时按照常规的线段树的update操作，这时候还应该更新rt子节点的sum[]值，而Lazy思想恰恰是暂时不更新rt子节点的sum[]值，到此就return，直到下次需要用到rt子节点的值的时候才去更新，这样避免许多可能无用的操作，从而节省时间 。

代码如下:

#include<stdio.h>#include<string.h>struct stu{int l,r;int mid(){return (l+r)>>1;}};stu node[1000000];__int64 sum[1000000];__int64 add[1000000];void PUTDOWN(int rt,int m){if(add[rt]){add[rt<<1]+=add[rt];add[rt<<1|1]+=add[rt];sum[rt<<1]+=add[rt]*(m-(m>>1));//算数运算符的优先级大于位运算符 ,竟然在这里错了. sum[rt<<1|1]+=add[rt]*(m>>1);add[rt]=0;}}void PUTUP(int rt){sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void build(int rt,int l,int r){node[rt].l=l;node[rt].r=r;add[rt]=0;if(l==r){scanf("%I64d",&sum[rt]);return;}int m=node[rt].mid();build(rt<<1,l,m);build(rt<<1|1,m+1,r);//2的倍数的二进制最后一位都为0       PUTUP(rt);//一开始每个结点的和 }void updata(int rt,int l,int r,int v){if(node[rt].l==l&&node[rt].r==r){add[rt]+=v;sum[rt]+=(__int64)v*(r-l+1);return;}PUTDOWN(rt,node[rt].r-node[rt].l+1);int m=node[rt].mid();if(r<=m)updata(rt<<1,l,r,v);else{if(l>m)updata(rt<<1|1,l,r,v);else {updata(rt<<1,l,m,v);updata(rt<<1|1,m+1,r,v);}}PUTUP(rt);}__int64 query(int rt ,int l,int r){if(node[rt].l==l&&node[rt].r==r)return sum[rt];PUTDOWN(rt,node[rt].r-node[rt].l+1);int m=node[rt].mid() ;if(r<=m) return query(rt<<1,l,r);else {if(l>m)return query(rt<<1|1,l,r);else return query(rt<<1,l,m)+query(rt<<1|1,m+1,r);}}int main(){int n,m,a,b,v;char c;while(scanf("%d%d",&n,&m)!=EOF){build(1,1,n);while(m--){getchar();scanf("%c%d%d",&c,&a,&b);if(c=='Q'){printf("%I64d\n",query(1,a,b));}else if(c=='C'){scanf("%d",&v);updata(1,a,b,v);}}}return 0;}