POJ 3468 线段树 成段更新 懒惰标记

A Simple Problem with Integers

Time Limit:5000MS   Memory Limit:131072K

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

Sample Output

4
55
9
15

Hint

The sums may exceed the range of 32-bit integers.

 

题解：本题也是线段树系列的模板题之一，要求的是成段更新+懒惰标记。PS：原题的说明有问题，上面说的“C a b c”中的c的范围其实在int32之外，需要使用long long，否则定是WA，真心坑爹，连WA多次，还是在discuss中看到的原因。

 

稍微讲解下代码中的一些细节: step<<1 与 step<<1|1，意思分别是step*2 和step*+1，具体为什么，可以回去复习一下位运算

 

转自 http://www.cnblogs.com/kevince/p/3888608.html

 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<map>
 
 #define N 100010
 #define M 15
 #define mod 1000000007
 #define mod2 100000000
 #define ll long long
 #define maxi(a,b) (a)>(b)? (a) : (b)
 #define mini(a,b) (a)<(b)? (a) : (b)
 
 using namespace std;
 
 int n,q;
 int x[N];
 char ch;
 int a,b,c;
 
 struct node
 {
     ll mark,sum;
 }tree[*N];
 
 void update(int l,int r,int i)
 {
     if(!tree[i].mark) return;
     int mid=(l+r)/;
     tree[*i].sum+=tree[i].mark*(ll)(mid-l+);
     tree[*i+].sum+=tree[i].mark*(ll)(r-mid);
     tree[*i].mark+=tree[i].mark;
     tree[*i+].mark+=tree[i].mark;
     tree[i].mark=;
 
 }
 
 ll query(int tl,int tr,int l,int r,int i)
 {
     if(tl>r || tr<l)
         return ;
     if(tl<=l && r<=tr)
         return tree[i].sum;
     update(l,r,i);
     int mid=(l+r)/;
 
     return query(tl,tr,l,mid,*i)+query(tl,tr,mid+,r,*i+);
 }
 
 void addd(int tl,int tr,int l,int r,int i,int val)
 {
     if(tl>r || tr<l)
         return;
     if(tl<=l && tr>=r){
         tree[i].sum+=val*(ll)(r-l+);
         tree[i].mark+=val;
         return;
     }
     update(l,r,i);
     int mid=(l+r)/;
     addd(tl,tr,l,mid,*i,val);
     addd(tl,tr,mid+,r,*i+,val);
     tree[i].sum=tree[*i].sum+tree[*i+].sum;
 }
 
 void build_tree(int l,int r,int i)
 {
     if(l==r){
         tree[i].sum=x[l];
         return;
     }
     int mid=(l+r)/;
     build_tree(l,mid,*i);
     build_tree(mid+,r,*i+);
     tree[i].sum=tree[*i].sum+tree[*i+].sum;
 }
 
 int main()
 {
     int i;
    // freopen("data.in","r",stdin);
     //scanf("%d",&T);
     //for(int cnt=1;cnt<=T;cnt++)
     //while(T--)
     while(scanf("%d%d",&n,&q)!=EOF)
     {
        // printf(" %d %d \n",n,q);
        for(i=;i<=n;i++) scanf("%d",&x[i]);
       // printf(" 1\n");
        memset(tree,,sizeof(tree));
        build_tree(,n,);
 
       //  printf(" 2\n");
        getchar();
        while(q--){
           //  printf(" 3\n");
             scanf("%c",&ch);
           //  printf(" %c\n",ch);
             if(ch=='C'){
                 //printf(" 31\n");
                 scanf("%d%d%d",&a,&b,&c);
                 getchar();
                 addd(a,b,,n,,c);
 
             }
             else{
                   //  printf(" 32\n");
                 scanf("%d%d",&a,&b);
                 getchar();
                 ll ans=query(a,b,,n,);
                 printf("%I64d\n",ans);
             }
        }
     }
 
     return ;
 }