POJ.3468 A Simple Problem with Integers(线段树 区间更新 区间查询)

POJ.3468 A Simple Problem with Integers(线段树 区间更新 区间查询)

题意分析

注意一下懒惰标记,数据部分和更新时的数字都要是long long ,别的没什么大坑。

代码总览

#include <cstdio>
#include <cstring>
#include <algorithm>
#define nmax 200000
using namespace std;
struct Tree{
    int l,r;
    long long val;
    long long  lazy;
    int mid(){
        return (l+r)>>1;
    }
};
Tree tree[nmax<<2];
long long  num[nmax];
void PushUp(int rt)
{
    tree[rt].val = tree[rt<<1].val + tree[rt<<1|1].val;
}
void PushDown(int rt)
{
    if(tree[rt].lazy){
        tree[rt<<1].lazy += tree[rt].lazy;
        tree[rt<<1|1].lazy += tree[rt].lazy;
        tree[rt<<1].val += tree[rt].lazy * (tree[rt<<1].r - tree[rt<<1].l +1) ;
        tree[rt<<1|1].val +=tree[rt].lazy * (tree[rt<<1|1].r - tree[rt<<1|1].l +1);
        tree[rt].lazy = 0;
    }
}
void Build(int l, int r, int rt)
{
    tree[rt].l = l; tree[rt].r = r;
    tree[rt].val = tree[rt].lazy = 0;
    if(l == r){
        tree[rt].val = num[l];
        return;
    }
    Build(l,tree[rt].mid(),rt<<1);
    Build(tree[rt].mid()+1,r,rt<<1|1);
    PushUp(rt);
}
void UpdatePoint(long long  val, int pos, int rt)
{
    if(tree[rt].l == tree[rt].r){
        tree[rt].val+=val;
        return;
    }
    PushDown(rt);
    if(pos<= tree[rt].mid()) UpdatePoint(val,pos,rt<<1);
    else UpdatePoint(val,pos,rt<<1|1);
    PushUp(rt);
}
void UpdateInterval(long long val, int l, int r, int rt)
{
    if(tree[rt].l >r || tree[rt].r < l) return;
    if(tree[rt].l >= l && tree[rt].r <= r){
        tree[rt].val += val * (tree[rt].r - tree[rt].l + 1);
        tree[rt].lazy += val;
        return;
    }
    PushDown(rt);
    UpdateInterval(val,l,r,rt<<1) ;
    UpdateInterval(val,l,r,rt<<1|1);
    PushUp(rt);
}
long long Query(int l,int r,int rt)
{
    if(l>tree[rt].r || r<tree[rt].l) return 0;
    PushDown(rt);
    if(l <= tree[rt].l && tree[rt].r <= r) return tree[rt].val;
    else return Query(l,r,rt<<1) + Query(l,r,rt<<1|1);

}
int N,Q,a,b;
long long c;
char op;
int main()
{
    //freopen("in3468.txt","r",stdin);
    while(scanf("%d %d",&N,&Q) != EOF){
        for(int i = 1;i<=N;++i) scanf("%I64d",&num[i]);
        Build(1,N,1);
        for(int i = 0;i<Q;++i){
            scanf(" %c %d %d",&op,&a,&b);
            if(op == 'C'){
                scanf("%I64d",&c);
                UpdateInterval(c,a,b,1);
            }else{
                printf("%I64d
",Query(a,b,1));
            }
        }
    }
    return 0;
}
原文地址:https://www.cnblogs.com/pengwill/p/7367043.html