poj 3468

来源：互联网发布：php和java的区别知乎编辑：程序博客网时间：2024/06/05 18:23

一道入门级的区间更新的线段树。

区间更新与单点更新相比，有一个lazy优化。就是一旦找到那个可以更新的区间，将那个区间更新完之后就不再继续更新，只是做一个lzy标记，等到下一次更新或查询时到更低层次的节点时，再顺便更新。在更新的过程中要注意是直接赋值还是加上这个值。还有题目说了这道题要用64位整型。

再来谈一下我的代码需要注意的地方，首先是从0开始计数，用开左闭右开区间表示范围，所以在查询和更新时一定要注意下标的对应，用[a-1,b)区间代表要查询或更新的[a,b]区间。

代码（C++）：

#include <iostream>#include <cstdio>#include <cstring>#define MAX 100009using namespace std;__int64 sum[MAX<<2],res;int num[MAX],lzy[MAX<<2];void build(int v,int h,int r){    int m,chl,chr;    if(r-h==1)    {        sum[v]=num[h];        return;    }    m=(h+r)>>1;    chl=v<<1|1;    chr=(v<<1)+2;    build(chl,h,m);    build(chr,m,r);    sum[v]=sum[chl]+sum[chr];}void init(int n){    memset(lzy,0,sizeof(lzy));    build(0,0,n);}void push_down(int v,int h,int r){    int m,chl,chr;    m=(h+r)>>1;    chl=v<<1|1;    chr=(v<<1)+2;    if(lzy[v]!=0)    {        sum[chl]+=(__int64)(m-h)*lzy[v];        sum[chr]+=(__int64)(r-m)*lzy[v];        lzy[chl]+=lzy[v];        lzy[chr]+=lzy[v];        lzy[v]=0;    }}void update(int a,int b,int c,int v,int h,int r){    int m,chl,chr;    if(a>=r||b<=h) return;    if(a<=h&&b>=r)    {        sum[v]+=(__int64)(r-h)*c;        lzy[v]+=c;        return;    }    m=(h+r)>>1;    chl=v<<1|1;    chr=(v<<1)+2;    push_down(v,h,r);    update(a,b,c,chl,h,m);    update(a,b,c,chr,m,r);    sum[v]=sum[chl]+sum[chr];}void query(int a,int b,int v,int h,int r){    int m,chr,chl;    if(a>=r||b<=h) return;    if(a<=h&&b>=r)    {        res+=sum[v];        return;    }    m=(r+h)>>1;    chl=v<<1|1;    chr=(v<<1)+2;    push_down(v,h,r);    query(a,b,chl,h,m);    query(a,b,chr,m,r);}int main(){    //freopen("in.txt","r",stdin);    int n,op,i,a,b,c;    char ch;    while(scanf("%d %d",&n,&op)!=EOF)    {        for(i=0;i<n;i++) scanf("%d",&num[i]);        init(n);        for(i=0;i<op;i++)        {            getchar();            scanf("%c",&ch);            if(ch=='Q')            {                scanf("%d %d",&a,&b);                res=0;                query(a-1,b,0,0,n);                printf("%I64d\n",res);            }else{                scanf("%d %d %d",&a,&b,&c);                update(a-1,b,c,0,0,n);            }        }    }    return 0;}

题目（http://poj.org/problem?id=3468）：

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072K 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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

Hint

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

0 0