POJ 3468 A Simple Problem with Integers（线段树）

题目链接：http://poj.org/problem?id=3468

题目思路：区间修改，区间查询，线段树模板题

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const double DINF = 0xffffffffffff;
const int mod = 1e9+7;

const int N = 1e6+5;

ll  a[N],n;  //输入数据,构造线段树数组

struct node{
    int lft,rht;
    ll sum;    //区间和
    ll lazy;   //延迟标记,减小时间复杂度
}segTree[N*4];  //需要开4倍

void pushUp(int id){    //区间合并,上放
    segTree[id].sum = segTree[id*2].sum+segTree[id*2+1].sum;
}

void pushDown(int id){
    if(segTree[id].lazy){   //区间修改过,需要下放
        //在原来的值上加上val
        segTree[id*2].sum += (segTree[id*2].rht-segTree[id*2].lft+1)*segTree[id].lazy;
        segTree[id*2+1].sum += (segTree[id*2+1].rht-segTree[id*2+1].lft+1)*segTree[id].lazy;
        segTree[id*2].lazy += segTree[id].lazy;
        segTree[id*2+1].lazy += segTree[id].lazy;
        segTree[id].lazy = 0;
    }
}

void build(int id,int l,int r){
    segTree[id].lft = l, segTree[id].rht = r;
    segTree[id].lazy = 0, segTree[id].sum = 0;   //开始一定要清0
    if(l == r){            //到达叶子节点,不继续建树
        segTree[id].sum = a[l];
    }
    else{                  //否则继续建树
        int mid = (l+r)>>1;
        build(id*2,l,mid);
        build(id*2+1,mid+1,r);
        pushUp(id);
    }
}

void upDate(int id,int l,int r,int val){    //更新l~r区间,加val,或减val(就传-val),或改成val
    if(l<=segTree[id].lft&&r>=segTree[id].rht){
        /*1.把原来的值加上val,因为该区间有segTree[id].rht-segTree[index].lft+1
        个数，所以区间和 以及 最值为：*/
        segTree[id].sum += (segTree[id].rht-segTree[id].lft+1)*val;
        segTree[id].lazy += val;  //延迟标记
    }
    else{
        pushDown(id);   //区间下放
        int mid = (segTree[id].lft+segTree[id].rht)>>1;
        if(r <= mid)
            upDate(id*2,l,r,val);
        else if(l>=mid+1)
            upDate(id*2+1,l,r,val);
        else{
            upDate(id*2,l,r,val);
            upDate(id*2+1,l,r,val);
        }
        pushUp(id);
    }
}

ll query(int id,int l,int r){  //查询l~r的值
    if(l<=segTree[id].lft&&r>=segTree[id].rht){  //该区间包含在查询区间内，可直接返回
        return segTree[id].sum;
    }
    pushDown(id);  //区间下放
    int mid = (segTree[id].lft+segTree[id].rht)>>1;
    ll ans = 0;
    if(r<=mid){  //只用管左子树
        ans += query(id*2,l,r);
    }
    else if(l>=mid+1){  //只用管右子树
        ans+=query(id*2+1,l,r);
    }
    else{
        ans += query(id*2,l,r)+query(id*2+1,l,r);
    }
    return ans;
    //return maxx;
    //return minn;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    ll t,l,r,x,m;
    char op;
    while(~scanf("%lld%lld",&n,&m)){
        for(int i=1;i<=n;i++)
            scanf("%lld",&a[i]);
        build(1,1,n);
        while(m--){
            scanf(" %c",&op);
            if(op=='Q')
                scanf("%lld%lld",&l,&r),printf("%lld\n",query(1,l,r));
            else
                scanf("%lld%lld%lld",&l,&r,&x),upDate(1,l,r,x);
        }
    }
    return 0;
}