Submission #527731

#	Time	Username	Problem	Language	Result	Execution time	Memory
527731		qwerasdfzxcl	식물 비교 (IOI20_plants)	C++14	44 / 100	945 ms	48496 KiB

plants

#include "plants.h"
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int INF = 1e9;
int n, a[200200], b[200200], idxa[200200], idxb[200200];
struct Seg2{
    pair<int, int> tree[800800];
    int lazy[800800];
    void init(int i, int l, int r){
        lazy[i] = 0;
        if (l==r) {tree[i] = {INF, l}; return;}

        int m = (l+r)>>1;
        init(i<<1, l, m); init(i<<1|1, m+1, r);
        tree[i] = min(tree[i<<1], tree[i<<1|1]);
    }
    void propagate(int i, int l, int r){
        tree[i].first += lazy[i];
        if (l!=r) lazy[i<<1] += lazy[i], lazy[i<<1|1] += lazy[i];
        lazy[i] = 0;
    }
    void update(int i, int l, int r, int s, int e, int x){
        propagate(i, l, r);
        if (r<s || e<l) return;
        if (s<=l && r<=e){
            lazy[i] += x;
            propagate(i, l, r);
            return;
        }
        int m = (l+r)>>1;
        update(i<<1, l, m, s, e, x); update(i<<1|1, m+1, r, s, e, x);
        tree[i] = min(tree[i<<1], tree[i<<1|1]);
    }
    void update2(int i, int l, int r, int s, int x){
        propagate(i, l, r);
        if (r<s || s<l) return;
        if (l==r) {tree[i].second = x; return;}
        int m = (l+r)>>1;
        update2(i<<1, l, m, s, x); update2(i<<1|1, m+1, r, s, x);
        tree[i] = min(tree[i<<1], tree[i<<1|1]);
    }
    void update(int l, int r, int x){
        if (r<=n) update(1, 1, n, l, r, x);
        else{
            update(1, 1, n, l, n, x);
            update(1, 1, n, 1, r-n, x);
        }
    }
}tree2;
struct Seg1{
    int tree[800800], lazy[800800];
    void init(int i, int l, int r, vector<int> &vec){
        lazy[i] = 0;
        if (l==r) {tree[i] = vec[l-1]; return;}

        int m = (l+r)>>1;
        init(i<<1, l, m, vec); init(i<<1|1, m+1, r, vec);
        tree[i] = max(tree[i<<1], tree[i<<1|1]);
    }
    void propagate(int i, int l, int r){
        tree[i] += lazy[i];
        if (l!=r) lazy[i<<1] += lazy[i], lazy[i<<1|1] += lazy[i];
        lazy[i] = 0;
    }
    void update(int i, int l, int r, int s, int e, int x){
        propagate(i, l, r);
        if (r<s || e<l) return;
        if (s<=l && r<=e){
            lazy[i] += x;
            propagate(i, l, r);
            return;
        }

        int m = (l+r)>>1;
        update(i<<1, l, m, s, e, x); update(i<<1|1, m+1, r, s, e, x);
        tree[i] = max(tree[i<<1], tree[i<<1|1]);
    }
    void find(int i, int l, int r, int x){
        propagate(i, l, r);
        if (tree[i]<x) return;
        if (l==r){
            assert(tree[i]==x);
            //printf("YES: %d\n", l);
            tree2.update(l, l, -INF);
            tree2.update(l+1, l+x, 1);

            update(1, 1, n, l, l, -INF);
            return;
        }

        int m = (l+r)>>1;
        find(i<<1, l, m, x); find(i<<1|1, m+1, r, x);
        tree[i] = max(tree[i<<1], tree[i<<1|1]);
    }
    void update(int s, int e, int x){
        if (s>0) update(1, 1, n, s, e, x);
        else{
            update(1, 1, n, s+n, n, x);
            update(1, 1, n, 1, e, x);
        }
    }
}tree1;

struct Seg3{
    pair<int, int> tree[400400];
    int sz;
    void init(int n, int *a){
        sz = n+1;
        for (int i=sz;i<sz*2;i++) tree[i] = {a[i-sz], i-sz};
        for (int i=sz-1;i;i--) tree[i] = min(tree[i<<1], tree[i<<1|1]);
    }
    void update(int p, int x){
        for (tree[p+=sz].first=x;p>1;p>>=1) tree[p>>1] = min(tree[p], tree[p^1]);
    }
    pair<int, int> query(int l, int r){
        pair<int, int> ret = {INF, -1};
        for (l+=sz, r+=sz;l<r;l>>=1, r>>=1){
            if (l&1) ret = min(ret, tree[l++]);
            if (r&1) ret = min(ret, tree[--r]);
        }
        return ret;
    }
    int queryc(int l, int r){
        if (l<=0){
            auto p = min(query(l+n, sz), query(1, r+1));
            return p.second;
        }
        if (r>n){
            auto p = min(query(l, sz), query(1, r-n+1));
            return p.second;
        }
        return query(l, r+1).second;
    }
}tree3;

vector<int> r, adj[200200];
int k, ans1[200200], ans2[200200], cnt1, cnt2, in[200200];
bool visited1[200200], visited2[200200];

void dfs1(int s){
    visited1[s] = 1;
    for (auto &v:adj[s]) if (!visited1[v]) dfs1(v);
    ans1[s] = cnt1--;
}

bool comp(int x, int y){return ans1[x] < ans1[y];}

void dfs2(int s){
    visited2[s] = 1;
    sort(adj[s].begin(), adj[s].end(), comp);
    for (auto &v:adj[s]) if (!visited2[v]) dfs2(v);
    ans2[s] = cnt2--;
}

void init(int K, vector<int> R) {
    k = K, r = R;
    n = r.size();
    tree1.init(1, 1, n, r);
    tree2.init(1, 1, n);
    tree2.update2(1, 1, n, 1, INF+1);

    for (int i=1;i<=n;i++){
        tree1.find(1, 1, n, k-1);
        assert(tree2.tree[1].first==0);
        a[i] = tree2.tree[1].second;
        if (a[i]>=INF) a[i] -= INF;
        //printf("a[%d] = %d\n", i, a[i]);

        tree2.update(a[i], a[i], INF);
        tree2.update(a[i]+1, a[i]+k-1, -1);
        tree1.update(a[i]-k+1, a[i]-1, 1);
    }

    for (int i=1;i<=n;i++) idxa[a[i]] = i;

    //printf("idxa: ");
    //for (int i=1;i<=n;i++) printf("%d ", idxa[i]);
    //printf("\n");

    tree3.init(n, idxa);
    for (int i=1;i<=n;i++){
        int pos = a[i];
        int l = tree3.queryc(pos-k+1, pos-1), r = tree3.queryc(pos+1, pos+k-1);
        if (l!=-1){
            adj[pos].push_back(l);
            in[l]++;
        }
        if (r!=-1){
            adj[pos].push_back(r);
            in[r]++;
        }

        //printf("%d: %d %d\n", pos, l, r);

        tree3.update(pos, INF+1);
    }

    for (int i=1;i<=n;i++) if (!in[i]) adj[0].push_back(i);
    cnt1 = cnt2 = n;
    dfs1(0);
    dfs2(0);

    //for (int i=1;i<=n;i++) printf("%d ", ans1[i]);
    //printf("\n");
    //for (int i=1;i<=n;i++) printf("%d ", ans2[i]);
    //printf("\n");
}

int compare_plants(int x, int y) {
    ++x, ++y;

    if (ans1[x] < ans1[y] && ans2[x] < ans2[y]) return -1;
    if (ans1[y] < ans1[x] && ans2[y] < ans2[x]) return 1;
    return 0;
}

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