POJ 3468 A Simple Problem with Integers 线段树 区间更新

原题： http://poj.org/problem?id=3468

题目：

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072K

Total Submissions: 78207 Accepted: 24097

Case Time Limit: 2000MS

Description

You have N integers, A1, A2, … , AN. 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, … , AN. -1000000000 ≤ Ai ≤ 1000000000.

Each of the next Q lines represents an operation.

“C a b c” means adding c to each of Aa, Aa+1, … , Ab. -10000 ≤ c ≤ 10000.

“Q a b” means querying the sum of Aa, Aa+1, … , Ab.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5

1 2 3 4 5 6 7 8 9 10

Q 4 4

Q 1 10

Q 2 4

C 3 6 3

Q 2 4

Sample Output

4

55

9

15

Hint

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

思路：

直接套模版。

注意数据会int溢出。

代码：

#include <iostream>
#include"cstdio"
#include"string.h"
using namespace std;
typedef long long int lint;
const int N= 100005;
struct node
{
int left;
int right;
lint add;
lint sum;
} tree[N*4];

int n,m;
lint a
;

void init()
{
memset(tree,0,sizeof(tree));
memset(a,0,sizeof(a));
}

//将下面更新的值返回上一层
void pushup(int id)
{
tree[id].sum=tree[id*2].sum+tree[id*2+1].sum;
}

//将标记区间向下移动
void pushdown(int id)
{
tree[id*2].add=tree[id*2].add+tree[id].add;
tree[id*2].sum=tree[id*2].sum+(tree[id*2].right-tree[id*2].left+1)*tree[id].add;
tree[id*2+1].add=tree[id*2+1].add+tree[id].add;
tree[id*2+1].sum=tree[id*2+1].sum+(tree[id*2+1].right-tree[id*2+1].left+1)*tree[id].add;
tree[id].add=0;
}

//建树的时候每次建完下一层要向上层赋值
void build(int id,int l,int r)
{
tree[id].left=l;
tree[id].right=r;
tree[id].add=0;
tree[id].sum=0;
if(l==r)
{
tree[id].sum=a[l];
return;
}
int mid=(l+r)/2;
build(id*2,l,mid);
build(id*2+1,mid+1,r);
pushup(id);
}

void add(int id,int l,int r,lint val)
{
//如果当前区间完全被包含，只更新当前区间的总和
//并对该区间做增加标记
if(tree[id].left>=l&&tree[id].right<=r)
{
tree[id].add=tree[id].add+val;
tree[id].sum=tree[id].sum+(tree[id].right-tree[id].left+1)*val;
return ;
}
//如果当前区间没有被包含
if(tree[id].right<l||tree[id].left>r)   return ;
//如果当前区间部分被包含，标记下移
if(tree[id].add)    pushdown(id);
//更新左右子区间
add(id*2,l,r,val);
add(id*2+1,l,r,val);
//更新完再把结果返回上层
pushup(id);
}

lint ans;
void query(int id,int l,int r)
{
//查询区间在该区间外
if(tree[id].left>r||tree[id].right<l)   return ;
//该区间完全被包含
if(tree[id].left>=l&&tree[id].right<=r)
{
ans=ans+tree[id].sum;
return ;
}
//查询部分该区间，标记下移
if(tree[id].add)    pushdown(id);
//左右查找
int mid=(tree[id].left+tree[id].right)/2;
if(l<=mid)  query(id*2,l,r);
if(r>mid)   query(id*2+1,l,r);
}

int main()
{
//freopen("in.txt","r",stdin);
while(scanf("%d %d",&n,&m)!=EOF)
{
init();
char s[10];
for(int i=1; i<=n; i++)
scanf("%I64d",&a[i]);
build(1,1,n);
for(int i=1; i<=m; i++)
{
int a,b,c;
scanf("%s",s);
if(s[0]=='Q')
{
ans=0;
scanf("%d %d",&a,&b);
query(1,a,b);
printf("%I64d\n",ans);
}
if(s[0]=='C')
{
scanf("%d %d %d",&a,&b,&c);
add(1,a,b,c);
}
}
}
//fclose(stdin);
return 0;
}