POJ-3468A Simple Problem with Integers，线段数区间更新查询，代码打了无数次还是会出错~~

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A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072K 吐舌头

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.

Source

POJ Monthly--2007.11.25, Yang Yi

这个题无非就是更新某个区间然后查询，思路甚至比其它的线段树要简单，但写起来却每次都出错；

思路很简单，如果每次更新都遍历到叶节点，毫无疑问会超时，既然是区间更新，我们可以先将要增加的值存在包含本区间的父亲区间里，下次往子节点操作时再进行类似的操作，也就是加上父亲节点储存的要增加的值，看代码+注释：

#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<algorithm>#include<cmath>using namespace std;const int  N=100000+10;int n,m,s[N];struct node{    int l,r;    long long n,add;//注意数据范围；} a[N<<2];void build(int l,int r,int k)//构建就没啥好讲的；{    a[k].l=l,a[k].r=r,a[k].add=0;    if(l==r)    {        a[k].n=s[l];        return ;    }    int mid=(l+r)/2;    build(l,mid,2*k);    build(mid+1,r,2*k+1);    a[k].n=a[k*2].n+a[k*2+1].n;}void update(int l,int r,int c,int k){    if(l<=a[k].l&&a[k].r<=r)    {        a[k].n+=(a[k].r-a[k].l+1)*c;//满足条件的话，这个区间每个元素都要加上c;        a[k].add+=c;将增加的值储存起来，下次更新操作时再往子节点更新；        return ;    }    if(a[k].add)//如果父亲节点增加值还在，那么子节点也要进行更新操作；    {        a[k*2].add+=a[k].add;        a[k*2+1].add+=a[k].add;        a[k*2].n+=(a[k*2].r-a[k*2].l+1)*a[k].add;        a[k*2+1].n+=(a[k*2+1].r-a[k*2+1].l+1)*a[k].add;        a[k].add=0;    }    int mid=(a[k].l+a[k].r)/2;    if(l<=mid) update(l,r,c,2*k);    if(r>mid) update(l,r,c,2*k+1);    a[k].n=a[k*2].n+a[k*2+1].n;回溯；}long long query(int l,int r,int k){    if(a[k].l==l&&a[k].r==r)        return a[k].n;    if(a[k].add)    {        a[k*2].add+=a[k].add;        a[k*2+1].add+=a[k].add;        a[k*2].n+=(a[k*2].r-a[k*2].l+1)*a[k].add;        a[k*2+1].n+=(a[k*2+1].r-a[k*2+1].l+1)*a[k].add;        a[k].add=0;    }    int mid=(a[k].l+a[k].r)/2;    if(l>mid) return query(l,r,2*k+1);    if(r<=mid) return query(l,r,2*k);    return query(l,mid,2*k)+query(mid+1,r,2*k+1);}int main(){    while(~scanf("%d%d",&n,&m))    {        memset(a,0,sizeof(a));        for(int i=1; i<=n; i++)            scanf("%d",&s[i]);        build(1,n,1);        while(m--)        {            int a,b,c;            getchar();            char o=getchar();            if(o=='Q')            {                scanf("%d%d",&a,&b);                printf("%I64d\n",query(a,b,1));            }            else            {                scanf("%d%d%d",&a,&b,&c);                update(a,b,c,1);            }        }    }    return 0;}

其实写出来发现也没什么改变，只不过关键在于更新和查询的时候往子节点所进行的操作；

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