Submission #304936

#	Time	Username	Problem	Language	Result	Execution time	Memory
304936		baluteshih	Comparing Plants (IOI20_plants)	C++14	0 / 100	1 ms	416 KiB

plants

#include "plants.h"
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define X first
#define Y second
#define ALL(v) v.begin(),v.end()
#define SZ(a) ((int)a.size())
#define pb push_back

int seg1[800005],seg2[800005],lazy1[800005],lazy2[800005];
int layer[400005],L[200005],R[200005];

void build(int l,int r,int rt,const vector<int> &v,int *seg)
{
    if(l==r)
        return seg[rt]=v[l],void();
    int mid=(l+r)>>1;
    build(l,mid,rt<<1,v,seg),build(mid+1,r,rt<<1|1,v,seg);
    seg[rt]=min(seg[rt<<1],seg[rt<<1|1]);
}

void down(int rt,int *seg,int *lazy)
{
    if(lazy[rt])
    {
        seg[rt<<1]+=lazy[rt],lazy[rt<<1]+=lazy[rt];
        seg[rt<<1|1]+=lazy[rt],lazy[rt<<1|1]+=lazy[rt];
        lazy[rt]=0;
    }
}

void modify(int L,int R,int l,int r,int rt,int val,int *seg,int *lazy)
{
    if(l!=r) down(rt,seg,lazy);
    if(L<=l&&R>=r)
        return lazy[rt]+=val,seg[rt]+=val,void();
    int mid=(l+r)>>1;
    if(L<=mid)
        modify(L,R,l,mid,rt<<1,val,seg,lazy);
    if(R>mid)
        modify(L,R,mid+1,r,rt<<1|1,val,seg,lazy);
    seg[rt]=min(seg[rt<<1],seg[rt<<1|1]);
}

int query(int l,int r,int rt,int *seg,int *lazy)
{
    if(l!=r) down(rt,seg,lazy);
    if(l==r)
        return l;
    int mid=(l+r)>>1;
    if(seg[rt<<1]<=seg[rt<<1|1])
        return query(l,mid,rt<<1,seg,lazy);
    return query(mid+1,r,rt<<1|1,seg,lazy);
}

void dfs(int l,int r,int rt,int *seg,int *lazy,vector<int> &rst)
{
    if(l!=r) down(rt,seg,lazy);
    if(l==r)
        return rst.pb(l);
    int mid=(l+r)>>1;
    if(seg[rt<<1]==0)
        dfs(l,mid,rt<<1,seg,lazy,rst);
    if(seg[rt<<1|1]==0)
        dfs(mid+1,r,rt<<1|1,seg,lazy,rst);
}

const int INF=1e9;
int arr[200005],K,n;

void init(int k, vector<int> r)
{
    n=SZ(r),K=k;
	build(0,n-1,1,r,seg1),build(0,n-1,1,vector<int>(n,INF),seg2);
    for(int i=0,ly=0;i<n;++ly)
    {
        while(seg1[1]==0)
        {
            int u=query(0,n-1,1,seg1,lazy1);
            modify(u,u,0,n-1,1,INF,seg1,lazy1),modify(u,u,0,n-1,1,-INF,seg2,lazy2);
            if(u+k<=n)
                modify(u+1,u+k-1,0,n-1,1,1,seg2,lazy2);
            else
            {
                if(u<n-1)
                    modify(u+1,n-1,0,n-1,1,1,seg2,lazy2);
                modify(0,k-n+u-1,0,n-1,1,1,seg2,lazy2);
            }
        }
        vector<int> rst;
        dfs(0,n-1,1,seg2,lazy2,rst);
        for(int u:rst)
        {
            if(u>=k-1)
                modify(u-k+1,u,0,n-1,1,-1,seg1,lazy1);
            else
                modify(0,u,0,n-1,1,-1,seg1,lazy1),modify(n-(k-u-1),n-1,0,n-1,1,-1,seg1,lazy1);
            if(u+k<=n)
                modify(u+1,u+k-1,0,n-1,1,-1,seg2,lazy2);
            else
            {
                if(u<n-1)
                    modify(u+1,n-1,0,n-1,1,-1,seg2,lazy2);
                modify(0,k-n+u-1,0,n-1,1,-1,seg2,lazy2);
            }
            modify(u,u,0,n-1,1,INF,seg2,lazy2);
            layer[u]=ly;
        }
        i+=SZ(rst);
    }
    /*for(int i=0;i<n;++i)
        cout << layer[i] << " \n"[i+1==n];*/
    for(int i=0;i<n;++i)
        layer[i+n]=layer[i];
    vector<int> st;
    for(int i=0;i<n;++i)
    {
        while(!st.empty()&&layer[i]<layer[st.back()])
            st.pop_back();
        st.pb(i);
    }
    for(int i=n;i<2*n;++i)
    {
        while(!st.empty()&&layer[i]<layer[st.back()])
            st.pop_back();
        L[i-n]=st.back()-n,st.pb(i);
    }
    st.clear();
    for(int i=2*n-1;i>=n;--i)
    {
        while(!st.empty()&&layer[i]<layer[st.back()])
            st.pop_back();
        st.pb(i);
    }
    for(int i=n-1;i>=0;--i)
    {
        while(!st.empty()&&layer[i]<layer[st.back()])
            st.pop_back();
        R[i]=st.back(),st.pb(i);
    }
}

int compare_plants(int x, int y)
{
    if(layer[x]==layer[y]) return 0;
    int rt=1;
    if(layer[x]>layer[y])
        swap(x,y),rt=-1;
    if((L[x]<=y&&y<=R[x])||(L[x]<=y-n&&y-n<=R[x])||(L[x]<=y+n&&y+n<=R[x]))
        return rt;
    return 0;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 14
• Subtask 3: 0 / 13
• Subtask 4: 0 / 17
• Subtask 5: 0 / 11
• Subtask 6: 0 / 15
• Subtask 7: 0 / 25