Vijos1066 Weak Trenches (Line Segment Tree)

topic analysis

# To fill the pit, how can
my line segment tree run slower than violence!
It should be a pot that didn't write lazy.

Code overview

segment tree

#include<bits/stdc++.h>
using namespace std;
const int nmax = 50000;
const int INF = 0x3f3f3f3f;
typedef struct{
    int l, r, v;
    int mid(){return ((l+r)>>1);}
}Tree;
Tree tree[nmax<<1];
typedef struct{
    int x, y;
}Node;
Node node[15005];
int ans[15005];
int n;
inline bool cmp(const Node & n1,const  Node & n2){
    if(n1.x == n2.x){
        if(n1.y <= n2.y) return true;
        else return false;
    }else return n1.x < n2.x;
}
void build(int l, int r, int rt){
    tree[rt].l = l; tree[rt].r = r;
    tree[rt].v = 0;
    if(tree[rt].l == tree[rt].r) return;
    build(l,tree[rt].mid(),rt<<1);
    build(tree[rt].mid()+1,r,rt<<1|1);
}
void update(int l, int r, int rt){
    if(tree[rt].l == tree[rt].r){
        tree[rt].v ++;
        return;
    }
    if(l <= tree[rt].mid()) update(l,r,rt<<1);
    if(r > tree[rt].mid()) update(l,r,rt<<1|1);
}
int query(int x ,int rt){
    if(tree[rt].l == tree[rt].r) return tree[rt].v;
    if( x <= tree[rt].mid()) return query(x,rt<<1);
    else return query(x,rt<<1|1);
}
int main(){
    scanf("%d",&n);
    for(int i = 0;i<n;++i) scanf("%d %d",&node[i].x, &node[i].y);
    sort(node,node+n,cmp);
    build(1,32001,1);
    for(int i = 0;i<n;++i){
        ans[query(node[i].y,1)]++;
        update(node[i].y,32001,1);
    }
    for(int i = 0;i<n;++i){
        printf("%d\n",ans[i]);
    }
    return 0;
}

Violence

#include<bits/stdc++.h>
using namespace std;
struct node{
    int x,y;
}a[15005];
int n;
int ans[15005];
int main(){
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>a[i].x>>a[i].y;
    for(int i=1;i<=n;i++){
        int tot=0;
        for(int j=1;j<=n;j++){
            if(i==j)
                continue;
            else
                if(a[j].x<=a[i].x&&a[j].y<=a[i].y)
                    tot++;
        }
        ans[tot]++;
    }
    for(int i=0;i<n;i++)
        cout<<ans[i]<<endl;
    return 0;
}