POJ 3468 A Simple Problemwith Integers(线段树:区间add,区间查询)

POJ 3468 A Simple Problemwith Integers(线段树:区间add,区间查询)

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

题意：

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.

分析：

        基本模板题目。下面代码中如果一个节点同时有addv值和sum值，那么这个节点控制元素的和只算sum，因为addv值早就已经被加了一遍了。其实我后来觉得应该两者都考虑，这样更符合思维习惯。

下面给出另一个延迟更新版的线段树区间add和区间求和的模板。

//POJ 3468 区间add，区间查询#include<cstdio>#include<cstring>#include<algorithm>using namespace std;//每当有add加到i节点上，不会去更新i节点的sum.//也就是说如果要查询区间[1,n]的sum值，既要考虑sum[i]的值，也要考虑add[i]的值const int MAXN=100000+100;typedef long long LL;#define lson i*2,l,m#define rson i*2+1,m+1,rint sum[MAXN*4];int addv[MAXN*4];//i节点收集下面节点的信息void PushUp(int i,int num){    sum[i]=sum[i*2]+sum[i*2+1]+addv[i*2]*(num+1)/2+addv[i*2+1]*(num-(num+1)/2);}//将i节点的addv压下去，且更新sum[i]void PushDown(int i,int num){    if(addv[i])    {        sum[i] += addv[i]*num;        addv[i*2]+=addv[i]; addv[i*2+1]+=addv[i];        addv[i]=0;    }}//sum[i]收集子节点的信息//只在build线段树时会用void PushUp(int i){    sum[i]=sum[i*2]+sum[i*2+1];}void build(int i,int l,int r){    addv[i]=0;    if(l==r)    {        scanf("%I64d",&sum[i]);        return ;    }    int m=(l+r)/2;    build(lson);    build(rson);    PushUp(i,r-l+1);}//给[ql,qr]区间加上add值void update(int ql,int qr,int add,int i,int l,int r){    if(ql<=l&&r<=qr)    {        addv[i]+=add;        //sum[i] += (LL)add*(r-l+1);        return ;    }    PushDown(i,r-l+1);    int m=(l+r)/2;    if(ql<=m) update(ql,qr,add,lson);    if(m<qr) update(ql,qr,add,rson);    PushUp(i,r-l+1);}//查询[ql,qr]区间的sum值int query(int ql,int qr,int i,int l,int r){    if(ql<=l&&r<=qr)    {        PushDown(i,r-l+1);        return sum[i];    }    PushDown(i,r-l+1);    int m=(l+r)/2;    int res=0;    if(ql<=m) res+=query(ql,qr,lson);    if(m<qr) res+=query(ql,qr,rson);    return res;}

AC代码：1797ms

//POJ 3468 区间add，区间查询#include<cstdio>#include<cstring>#include<algorithm>using namespace std;//每当有add加到i节点上，直接更新i节点的sum.//也就是说如果要查询区间[1,n]的sum值，直接sum[1]即可,不用再去考虑1的addv[1]值.const int MAXN=100000+100;typedef long long LL;#define lson i*2,l,m#define rson i*2+1,m+1,rLL sum[MAXN*4];LL addv[MAXN*4];void PushDown(int i,int num){    if(addv[i])    {        sum[i*2] +=addv[i]*(num-(num/2));        sum[i*2+1] +=addv[i]*(num/2);        addv[i*2] +=addv[i];        addv[i*2+1]+=addv[i];        addv[i]=0;    }}void PushUp(int i){    sum[i]=sum[i*2]+sum[i*2+1];}void build(int i,int l,int r){    addv[i]=0;    if(l==r)    {        scanf("%I64d",&sum[i]);        return ;    }    int m=(l+r)/2;    build(lson);    build(rson);    PushUp(i);}void update(int ql,int qr,int add,int i,int l,int r){    if(ql<=l&&r<=qr)    {        addv[i]+=add;        sum[i] += (LL)add*(r-l+1);        return ;    }    PushDown(i,r-l+1);    int m=(l+r)/2;    if(ql<=m) update(ql,qr,add,lson);    if(m<qr) update(ql,qr,add,rson);    PushUp(i);}LL query(int ql,int qr,int i,int l,int r){    if(ql<=l&&r<=qr)    {        return sum[i];    }    PushDown(i,r-l+1);    int m=(l+r)/2;    LL res=0;    if(ql<=m) res+=query(ql,qr,lson);    if(m<qr) res+=query(ql,qr,rson);    return res;}int main(){    int n,q;    while(scanf("%d%d",&n,&q)==2&&n&&q)    {        build(1,1,n);        while(q--)        {            char str[10];            scanf("%s",str);            int x,y,z;            if(str[0]=='Q')            {                scanf("%d%d",&x,&y);                printf("%I64d\n",query(x,y,1,1,n));            }            else            {                scanf("%d%d%d",&x,&y,&z);                update(x,y,z,1,1,n);            }        }    }    return 0;}