【poj3468】A Simple Problem with Integers
題面
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.
題解
線段樹模板
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 100010;
typedef long long LL;
LL a[maxn];
struct node{
int l, r;
LL val, addmark;
}sgt[maxn<<2];
void build(int p, int l, int r){
sgt[p].l = l, sgt[p].r = r;
if(l == r){
sgt[p].val = a[l];
}else{
int m = (l+r)/2;
build(p*2,l,m);
build(p*2+1,m+1,r);
sgt[p].val = sgt[p*2].val+sgt[p*2+1].val;
}
}
void pushdown(int p){
if(sgt[p].addmark != 0){
LL t = sgt[p].addmark;
sgt[p].addmark = 0;
sgt[p*2].addmark += t;
sgt[p*2+1].addmark += t;
sgt[p*2].val += (sgt[p*2].r-sgt[p*2].l+1)*t;
sgt[p*2+1].val += (sgt[p*2+1].r-sgt[p*2+1].l+1)*t;
}
}
void add(int p, int l, int r, LL v){
if(l <= sgt[p].l && sgt[p].r <= r){
sgt[p].val += (sgt[p].r-sgt[p].l+1)*v;
sgt[p].addmark += v;
return ;
}
pushdown(p);
int m = (sgt[p].l+sgt[p].r)/2;
if(l <= m)add(p*2,l,r,v);
if(r > m)add(p*2+1,l,r,v);
sgt[p].val = sgt[p*2].val+sgt[p*2+1].val;
}
LL query(int p, int l, int r){
if(l <= sgt[p].l && sgt[p].r <= r)return sgt[p].val;
pushdown(p); //pushdown
LL m = (sgt[p].l+sgt[p].r)/2, ans = 0;
if(l <= m)ans += query(p*2,l,r);
if(r > m)ans += query(p*2+1,l,r);
return ans;
}
int main(){
ios::sync_with_stdio(false);
int n, m;
cin>>n>>m;
for(int i = 1; i <= n; i++)cin>>a[i];
build(1,1,n);
for(int i = 1; i <= m; i++){
char ch; cin>>ch;
if(ch == 'Q'){
LL x, y; cin>>x>>y;
cout<<query(1,x,y)<<"\n";
}else{
LL x, y, z; cin>>x>>y>>z;
add(1,x,y,z);
}
}
return 0;
}
關於線段樹2*n空間表示法。
推薦部落格http://www.cppblog.com/MatoNo1/archive/2015/05/05/195857.html
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