poj3468A Simple Problem with Integers（线段树+成段更新） - c++编程基础

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poj3468A Simple Problem with Integers（线段树+成段更新）

2015-07-20 17:47:24 来源: 作者: 【大中小】浏览:1次

Tags：poj3468A Simple Problem with Integers 线段成段更新

题目链接：
huangjing
题意：
给n个数，然后有两种操作。
【1】Q a b 询问a到b区间的和。
【2】C a b c将区间a到b的值都增加c。
思路：
线段树成段更新的入门题目。。学会使用lazy即可。还需要注意的是，lazy的时候更改是累加，而不是直接修改。。有可能连续几次进行修改操作。。注意这一点就好了。。。

题目：

Language: A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 62629		Accepted: 19215
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

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi 代码：

#include
      
       
#include
       
         #include
        
          #include
         
           #include
           #include
           
             #include
            
              #include
             
               #include
              
                #define eps 1e-9 #define ll long long #define INF 0x3f3f3f3f using namespace std; const int maxn=100000+10; ll col[maxn*4]; ll tree[maxn*4]; int n,m; void push_up(int dex) { tree[dex]=tree[dex<<1]+tree[dex<<1|1]; } void push_down(int cnt,int dex) { if(col[dex]) { col[dex<<1]+=col[dex];//这里要累积起来 col[dex<<1|1]+=col[dex]; tree[dex<<1]+=(cnt-(cnt>>1))*col[dex]; tree[dex<<1|1]+=(cnt>>1)*col[dex]; col[dex]=0; } } void buildtree(int l,int r,int dex) { col[dex]=0; if(l==r) { scanf("%I64d",&tree[dex]); return; } int mid=(l+r)/2; buildtree(l,mid,dex<<1); buildtree(mid+1,r,dex<<1|1); push_up(dex); } void update(int L,int R,int l,int r,int dex,long long val) { if(L<=l&&R>=r) { tree[dex]+=(r-l+1)*val; col[dex]+=val; return; } push_down(r-l+1,dex); int mid=(l+r)/2; if(L<=mid) update(L,R,l,mid,dex<<1,val); if(R>mid) update(L,R,mid+1,r,dex<<1|1,val); push_up(dex); } long long Query(int L,int R,int l,int r,int dex) { if(L<=l&&R>=r) return tree[dex]; push_down(r-l+1,dex); int mid=(l+r)/2; if(R<=mid) return Query(L,R,l,mid,dex<<1); else if(L>mid) return Query(L,R,mid+1,r,dex<<1|1); else return Query(L,R,l,mid,dex<<1)+Query(L,R,mid+1,r,dex<<1|1); } int main() { char str[2]; int u,v; long long w; while(~scanf("%d%d",&n,&m)) { buildtree(1,n,1); while(m--) { scanf("%s",str); if(str[0]=='Q') { scanf("%d%d",&u,&v); printf("%I64d\n",Query(u,v,1,n,1)); } else { scanf("%d%d%I64d",&u,&v,&w); update(u,v,1,n,1,w); } } } return 0; }


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