“21天好习惯”第一期-14 线段树

21天零基础入门ACM

21天零基础入门ACM之 第14天

线段树

线段树

线段树就是对一些满足区间可加性的数组的特性进行预处理。

线段树的基本实现

操作：

void pushup(int k){
    tree[k]=tree[k<<1]+tree[k<<1|1];
}

建立：

void build(int k,int l,int r){
    if(l==r) {
        tree[k]=nums[l];
        return;
    }
    int mid=(l+r)>>1;
    build(k<<1,l,mid);
    build(k<<1|1,mid+1,r);
    pushup(k);
}

更新：

void updata(int pos,int val,int k,int l,int r){
    if(l==r) {
        nums[k]+=val;
        tree[k]+=val;
        return;
    }
    int mid = (l+r)>>1;
    if(pos<=mid) updata(pos,val,k<<1,l,mid);
    else  updata(pos,val,k<<1|1,mid+1,r);
    pushup(k);
}

查询：

ll query(int L,int R,int k,int l,int r){
    if(L<=l&&R>=r) return tree[k];
    ll ans=0;
    int mid = (l+r)>>1;
    if(L<=mid) ans+=query(L,R,k<<1,l,mid);
    if(R>mid) ans+=query(L,R,k<<1|1,mid+1,r);
    return ans;
}

例题：

例题1：
https://ac.nowcoder.com/acm/problem/15164


代码：

#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define sc(x) scanf("%d",x)
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    
ll qpow(ll a, ll b, ll mod) { ll ans = 1;  a%=mod; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); }    while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();    return s * w; }

const int mod=1e9+7;
const int N=1e5+7;

int n,m;
ll nums[N],tree[N<<2];

void pushup(int k){
    tree[k]=tree[k<<1]+tree[k<<1|1];
}

void build(int k,int l,int r){
    if(l==r) {
        tree[k]=nums[l];
        return;
    }
    int mid=(l+r)>>1;
    build(k<<1,l,mid);
    build(k<<1|1,mid+1,r);
    pushup(k);
}

void updata(int pos,int val,int k,int l,int r){
    if(l==r) {
        nums[k]+=val;
        tree[k]+=val;
        return;
    }
    int mid = (l+r)>>1;
    if(pos<=mid) updata(pos,val,k<<1,l,mid);
    else  updata(pos,val,k<<1|1,mid+1,r);
    pushup(k);
}

ll query(int L,int R,int k,int l,int r){
    if(L<=l&&R>=r) return tree[k];
    ll ans=0;
    int mid = (l+r)>>1;
    if(L<=mid) ans+=query(L,R,k<<1,l,mid);
    if(R>mid) ans+=query(L,R,k<<1|1,mid+1,r);
    return ans;
}
ll x,y,z;
int main(){
    cin>>n>>m;
    for(int i=1;i<=n;i++) cin>>nums[i];
    build(1,1 ,n );
    while(m--){
        cin>>x>>y>>z;
        if(x==1) updata(y,z,1,1,n);
        else cout<<query(y,z,1,1,n)<<endl;
    }
}

例题2：
https://ac.nowcoder.com/acm/problem/15163


代码：

#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define sc(x) scanf("%d",x)
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    
ll qpow(ll a, ll b, ll mod) { ll ans = 1;  a%=mod; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); }    while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();    return s * w; }

const int mod=1e9+7;
const int N=1e5+7;

int n,x;
int tree[N<<2];

void insert(int k,int l,int r,int x){
    ++tree[k];
    if(l==r) return;
    int mid=(l+r)>>1;
    if(x<=mid) insert(k<<1,l,mid,x);
    else insert(k<<1|1,mid+1,r,x);
}

ll query(int L,int R,int k,int l,int r){
    if(l>=L&&r<=R) return tree[k];
    ll ans=0;
    int mid=(l+r)>>1;
    if(L<=mid) ans+=query(L,R,k<<1,l,mid);
    if(mid<R) ans+=query(L,R, k<<1|1,mid+1, r);
    return ans;
}


int main(){
    n=read();
    ll ans=0;
    for(int i=0;i<n;i++){
        x=read();
        x++;
        ans+=query(x+1,N,1,1,N);
        insert(1,1,N,x);
    }
    cout<<ans<<endl;
}