A Simple Problem with Integers (POJ_3468) 线段树+区间更新
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 84588 | Accepted: 26203 | |
| 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.
题目大意：给出n个数，对这些树进行区间更新和查询操作
解题思路：利用线段树区间更新，注意区间更新时当节点区间全部包含于待更新区间时，不再对其下节点更新，而把更新值c存储在add中，当线段树的平衡被破坏时，再将值向下更新。
代码如下：
#include"iostream"
#include"cstdio"
#define ll __int64
using namespace std;
//线段树的最大节点数
const ll maxn=110000;
//线段数定义
struct Tree{
ll l,r,sum,add;
ll getmid(){ //返回线段的中点
return (l+r)>>1;
}
}tree[4*maxn];
//向上一层推
void pushup(ll x){
ll tmp=2*x;
tree[x].sum=tree[tmp].sum+tree[tmp+1].sum;
}
//向下一层推
void pushdown(ll x){
ll tmp=2*x;
tree[tmp].add+=tree[x].add;
tree[tmp+1].add+=tree[x].add;
tree[tmp].sum+=tree[x].add*(tree[tmp].r-tree[tmp].l+1);
tree[tmp+1].sum+=tree[x].add*(tree[tmp+1].r-tree[tmp+1].l+1);
tree[x].add=0;
}
//构建线段树
void Build(ll l,ll r,ll x){
tree[x].l=l;
tree[x].r=r;
tree[x].add=0;
if(tree[x].l==tree[x].r){ //叶子节点
scanf("%I64d",&tree[x].sum);
return ;
}
ll mid=tree[x].getmid();
ll tmp=2*x;
Build(l,mid,tmp); //向左子树递归
Build(mid+1,r,tmp+1); //向右子树递归
pushup(x); //在回溯时向上推
}
//区间更新
void Updata(ll l,ll r,ll x,ll c){ //所更新区间【l,r】，增加值c,当前节点x
if(tree[x].l>r||tree[x].r<l) //节点包含区间，不在更新区间内
return ;
if(l<=tree[x].l&&tree[x].r<=r){ //节点包含区间包含于更新区间
tree[x].add+=c;
tree[x].sum+=c*(tree[x].r-tree[x].l+1);
return ;
}
if(tree[x].add!=0)
pushdown(x); //在抵达新一轮更新区间前，将之前积攒的add向下更新
ll tmp=2*x;
Updata(l,r,tmp,c);
Updata(l,r,tmp+1,c);
pushup(x);
}
//区间查询
ll ans; //存储查询答案
void Query(ll l,ll r,ll x){ //【l,r】为当前查询区间，x为当前节点
if(tree[x].l>r||tree[x].r<l) //节点包含区间不在查询区间
return ;
if(l<=tree[x].l&&tree[x].r<=r){ //节点区间包含于查询区间
ans+=tree[x].sum;
return ;
}
if(tree[x].add!=0)
pushdown(x); //在抵达询问区间前，将之前积攒的add向下更新
ll mid=tree[x].getmid();
ll tmp=2*x;
if(r<=mid) //当前查询区间在节点的左半区间
Query(l,r,tmp);
else if(l>mid) //当前查询区间在节点的右半区间
Query(l,r,tmp+1);
else{ //查询区间分布在节点的左右区间
Query(l,mid,tmp);
Query(mid+1,r,tmp+1);
}
}
//主函数
int main(){
ll n,q;
while(scanf("%I64d%I64d",&n,&q)!=EOF){
Build(1,n,1);
char ch;
while(q--){
getchar();
ch=getchar();
ll x,y,c;
if(ch=='C'){
scanf("%I64d%I64d%I64d",&x,&y,&c);
Updata(x,y,1,c);
}
else{
scanf("%I64d%I64d",&x,&y);
ans=0;
Query(x,y,1);
printf("%I64d\n",ans);
}
}
}
return 0;
}