poj3468 线段树区间更新+区间查询

题目链接：

http://poj.org/problem?id=3468

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072K

Total Submissions: 149649 Accepted: 46438

Case Time Limit: 2000MS

Description

You have N integers, A1, A2, ... , AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

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.

Source

POJ Monthly--2007.11.25, Yang Yi
[Submit] [Go Back] [Status] [Discuss]

区间更新和和区间查询的模板

This is the code

#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<sstream>
#include<stack>
#include<string>
#include<set>
#include<vector>
using namespace std;
#define PI acos(-1.0)
#define pppp cout<<endl;
#define EPS 1e-8
#define LL long long
#define ULL unsigned long long     //1844674407370955161
#define INT_INF 0x3f3f3f3f      //1061109567
#define LL_INF 0x3f3f3f3f3f3f3f3f //4557430888798830399
// ios::sync_with_stdio(false);
// 那么cin, 就不能跟C的 scanf，sscanf, getchar, fgets之类的一起使用了。
const int dr[]={0, 0, -1, 1, -1, -1, 1, 1};
const int dc[]={-1, 1, 0, 0, -1, 1, -1, 1};const int MAXN=100005;
LL sum[MAXN<<2],lazy[MAXN<<2];
//sum[]模拟线段树维护区间和，lazy[]为懒惰标记void PushUp(int i)
{sum[i]=sum[i<<1]+sum[i<<1|1];
}void PushDown(int i,int ln,int rn)//ln表示左子树元素结点个数，rn表示右子树结点个数
{if (lazy[i]){lazy[i<<1]+=lazy[i];lazy[i<<1|1]+=lazy[i];sum[i<<1]+=lazy[i]*ln;sum[i<<1|1]+=lazy[i]*rn;lazy[i]=0;}
}
void Build(int l,int r,int i)//建树，Build(1,n,1);
{if (l==r){scanf("%lld",&sum[i]);return;}int mid=(l+r)>>1;Build(l,mid,i<<1);Build(mid+1,r,i<<1|1);PushUp(i);
}void Add(int L,int C,int l,int r,int i)//单点更新,Add(L,C,1,n,1);
{if (l==r){sum[i]+=C;return;}int mid=(l+r)>>1;PushDown(i,mid-l+1,r-mid); //若既有点更新又有区间更新，需要这句话if (L<=mid)Add(L,C,l,mid,i<<1);elseAdd(L,C,mid+1,r,i<<1|1);PushUp(i);
}void Update(int L,int R,int C,int l,int r,int i)//区间更新,Update(L,R,C,1,n,1);
{if (L<=l&&r<=R){sum[i]+=C*(r-l+1);lazy[i]+=C;return;}int mid=(l+r)>>1;PushDown(i,mid-l+1,r-mid);if (L<=mid)Update(L,R,C,l,mid,i<<1);if (R>mid)Update(L,R,C,mid+1,r,i<<1|1);PushUp(i);
}LL Query(int L,int R,int l,int r,int i)//区间查询,Query(L,R,1,n,1);
{if (L<=l&&r<=R)return sum[i];int mid=(l+r)>>1;PushDown(i,mid-l+1,r-mid);//若更新只有点更新，不需要这句LL ANS=0;if (L<=mid)ANS+=Query(L,R,l,mid,i<<1);if (R>mid)ANS+=Query(L,R,mid+1,r,i<<1|1);return ANS;
}int main()
{//freopen("D:\\chnegxubianji\\inORout\\in.txt", "r", stdin);//freopen("D:\\chnegxubianji\\inORout\\out.txt", "w", stdout);int n,q;scanf("%d%d",&n,&q);Build(1,n,1);while(q--){char str;scanf(" %c",&str);if(str=='Q'){int x,y;scanf("%d%d",&x,&y);printf("%lld\n",Query(x,y,1,n,1));}else{int x,y,w;scanf("%d%d%d",&x,&y,&w);Update(x,y,w,1,n,1);}}return 0;
}

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poj3468 线段树区间更新+区间查询

题目链接：

http://poj.org/problem?id=3468

This is the code

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