제출 #828959

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
828959		yeyso	밀림 점프 (APIO21_jumps)	C++14	23 / 100	1569 ms	309856 KiB

jumps

#include "jumps.h"
#include <bits/stdc++.h>
using namespace std;
// st0[u][z] = st0[st0[u][z-1]][z-1];
#include <vector>
vector<vector<int>> adj;
vector<vector<int>> radj;
int n = 0;
vector<vector<int>> st0;
vector<vector<int>> st1;
vector<int> visited;
vector<int> h;
// SEGMENT TREE
const int S_ = 262144 * 4 + 5;
vector<vector<int>> tree(S_, vector<int>());
// fast[l/r][node][number] = biggest element <= number in l/r subtree of node
// wait NVM
// For each element inside a segment tree node, we need the O(1) function for that node 
//vector<vector<map<int, int>>> fast(2, vector<map<int, int>>(262144 * 2 + 5, map<int, int>()));
// fast[l/r][node][index within that node] = biggest element <= node[index] in l/r subtree of node
vector<vector<vector<int>>> fast(2, vector<vector<int>>(262144 * 4 + 5, vector<int>()));
vector<vector<int>> RMQ;
/*
THE PLAN

To find the best tree in the range [A, B]
- Find the rightmost tree in the range that is taller than C, call it X using a 
Segtree that supports range rightmost threshold queries! 

Then a segtree that supports range upper bound queries!
A != B but C = D



A ... B 
Let X = rightmost tree between A and B that is taller than C
Now use range upper bound to find tallest tree between X and B that is shorter than C

And then do RMQ(C) for the tree we want to start at
Then we need to find the fastest way to get to (C -- D)
Solve the problem in reverse

- Then do the range upper bound query within the range [X, B]
*/
//vector<vector<vector<int>>> fast(2, vector<vector<int>>(262144 * 2 + 5, vector<int>()));
//int[2][262144 * 2 +5][] fast;
// For each node in the segment tree, we have O(log(n)) numbers
// n logn + n logn + n logn ... logn times

int fast_log(int x){
    return x ? __builtin_clzll(1) - __builtin_clzll(x) : -1;
}
void build(int j, int tl, int tr) {
    if (tl == tr) {
        tree[j] = {h[tl]};
        return;
    } else {
        int tm = (tl + tr) / 2;
        build(j*2, tl, tm);
        build(j*2+1, tm+1, tr);
        merge(tree[j * 2].begin(), tree[j * 2].end(), tree[j * 2 + 1].begin(), tree[j * 2 + 1].end(), back_inserter(tree[j]));
        int pl = 0;
        int pr = 0;
        
        for(int x = 0; x < tree[j].size(); x ++){
            if(tree[j * 2][pl] == tree[j][x]){
                fast[0][j].push_back(pl);
                if(x == 0){ fast[1][j].push_back(-1);
                } else { fast[1][j].push_back(fast[1][j][fast[1][j].size() - 1]); }
                pl += 1;
            } else {
                fast[1][j].push_back(pr);
                if(x == 0){ fast[0][j].push_back(-1);
                } else { fast[0][j].push_back(fast[0][j][fast[0][j].size() - 1]); }
                pr += 1;
            }
        }
 
        if(fast[1][j].size() == 0) fast[1][j].push_back(-1);
        if(fast[0][j].size() == 0) fast[0][j].push_back(-1);
    }
}
 
vector<int> query;
int rmq0(int l, int r){
    int i = fast_log(r - l + 1);
    return max(RMQ[i][l], RMQ[i][r - (1 << i) + 1]);
}
int qu = 0;
void rightmost(int threshold, int ql, int qr, int lbound, int rbound, int j = 1){
    int mid = (lbound + rbound) / 2;
    if(lbound == rbound){
        qu = max(qu, lbound);
        return;
    }
    int call = 0;
    if(rmq0(mid+1, rbound) >= threshold){
        if(ql <= rbound and qr >= mid + 1){
            call = 1;
            rightmost(threshold, ql, qr, mid+1, rbound, j * 2 + 1);
        }
    }
    if(rmq0(lbound, mid) >= threshold){
        if(!call){
            if(ql <= mid and qr >= lbound){
                rightmost(threshold, ql, qr, lbound, mid, j * 2);
            }
        }
    }
}
void rangeq(int bound, int ql, int qr, int lbound, int rbound, int j = 1){
    if(qr < lbound || rbound < ql){
        //return INT_MAX;
        // DONT DO ANYTHING
    } else if(ql <= lbound and rbound <= qr){
        if(bound >= 0){
            query.push_back(tree[j][bound]);
        }
    } else {
        int mid = (lbound + rbound) / 2;
        if(bound >= 0){
            rangeq(fast[0][j][bound], ql, qr, lbound, mid, j * 2);
            rangeq(fast[1][j][bound], ql, qr, mid+1, rbound, j * 2 + 1);
        }
    }
}
// Construct sparse tables
void dfs0(int u, int v){
    if(!visited[u]){
        visited[u] = 1;
        st0[u][0] = v;
        for(int z = 1; z <= ceil(log2(n)); z ++){
            st0[u][z] = st0[st0[u][z-1]][z-1];
        }
        for(int i = 0; i < radj[u].size(); i ++){
            int highedge = 1;
            for(int j = 0; j < adj[radj[u][i]].size(); j ++){
                if(h[adj[radj[u][i]][j]] > h[u]){
                    highedge = 0;
                }
            }
            if(highedge){
                dfs0(radj[u][i], u);
            }
        }
    }
}
void dfs1(int u, int v){
    if(!visited[u]){
        visited[u] = 1;
        st1[u][0] = v;
        for(int z = 1; z <= ceil(log2(n)); z ++){
            st1[u][z] = st1[st1[u][z-1]][z-1];
        }
        for(int i = 0; i < radj[u].size(); i ++){
            int lowedge = 1;
            for(int j = 0; j < adj[radj[u][i]].size(); j ++){
                if(h[adj[radj[u][i]][j]] < h[u]){
                    lowedge = 0;
                }
            }
            if(lowedge){
                dfs1(radj[u][i], u);
            }
        }
    }
}
    
int subtask1 = 1;
vector<int> htt;
void init(int N, vector<int> H) {
    n = N;
    h = H;
    stack<pair<int, int>> s;
    vector<vector<int>> adj0(N, vector<int>());
    int highest = 0;
    int idx = 0;
    vector<vector<int>> radj0(N, vector<int>());
    // Monotonic stack to construct adjacency list in O(n)
    for (int i = 0; i < N; i ++) {
        if(i > 0){
            if(H[i] != i + 1){
                subtask1 = 0;
            }
        }
        while (!s.empty() && s.top().first < H[i]){ s.pop(); }
        if (!s.empty()){
            adj0[i].push_back(s.top().second);
            radj0[s.top().second].push_back(i);
        }
        s.push({H[i], i});
        if(H[i] > highest){
            highest = H[i];
            idx = i;
        }
    }
    while(!s.empty()){ s.pop(); }
    for (int i = N - 1; i >= 0; i --) {
        while (!s.empty() && s.top().first < H[i]){ s.pop(); }
        if (!s.empty()){
            adj0[i].push_back(s.top().second);
            radj0[s.top().second].push_back(i);
        }
        s.push({H[i], i});
    }
    adj = adj0;
    radj = radj0;
    vector<vector<int>> st(N, vector<int>(ceil(log2(n)) + 1, 0));
    vector<vector<int>> rmqt(ceil(log2(n))+1, vector<int>(N, 0));
    st0 = st;
    st1 = st;
    RMQ = rmqt;
    for(int j = 0; j < n; j ++){
        RMQ[0][j] = H[j];
    }
    for(int i = 1; i <= ceil(log2(n)); i ++){
        for(int j = 0; j < n; j ++){
            RMQ[i][j] = max(RMQ[i-1][j], RMQ[i-1][j + (1 << (i - 1))]);
        }
    }
    vector<int> v(N, 0);
    visited = v;
    htt = v;
    for(int i = 0; i < H.size(); i ++){
        htt[H[i] - 1] = i;
    }
    dfs0(idx, idx);
    visited = v;
    dfs1(idx, idx);
    build(1, 0, N + 1);
}
 
int minimum_jumps(int A, int B, int C, int D){
    qu = -1;
    rightmost(h[C], A, B, 0, n);

    //return rmq0(A, B);
    int start = rmq0(max(qu, A), B);
    //return qu;
    
    if(C == D){
        int jumps = 0;
        int u = htt[start - 1];
        for (int i = ceil(log2(n)); i >= 0; --i) {
            if(h[st0[u][i]] < h[C]){
                u = st0[u][i];
                jumps += (1 << i);
            }
        }
        // If the first sparse table is all we need, we're done
        if(st0[u][0] == C){
            return jumps + 1;
        } else {
            // Otherwise start at the next sparse table
            for (int i = ceil(log2(n)); i >= 0; --i) {
                if(h[st1[u][i]] < h[C]){
                    u = st1[u][i];
                    jumps += (1 << i);
                }
            }
            
            if(st1[u][0] == C){
                return jumps + 1;
            }
        }
        return -1;
    }  /*{
        if(subtask1){
            return C - B;
        } else {
            queue<pair<int, int>> q;
            for(int i = A; i <= B; i ++){
                q.push({0, i});
            }
            int node = 0; int dist = 0;
            vector<int> vi(n, 0);
            while(!q.empty()){
                node = q.front().second;
                dist = q.front().first;
                //cout << node << "\n";
                q.pop();
                if(node >= C && node <= D){
                    return dist;
                    //break;
                }
                if(!vi[node]){
                    vi[node] = 1;
                    for(int i = 0; i < adj[node].size(); i ++){
                        q.push({dist + 1, adj[node][i]});
                    }
                }
            }
 
            return -1;
        }
    }*/
    return -1;
    
}
/*
The rightmost tree within range that is taller than 
g++ -std=gnu++17 -O2 -pipe -o jumps jumps.cpp stub.cpp

7 3
3 7 5 6 4 2 1
0 1 6 6
0 2 6 6
0 3 6 6
0 4 4 4
0 5 4 4
0 6 4 4
3 6 5 5
4 6 5 5
5 6 5 5
6 6 6 6
 
if(A == B and C == D){
        int jumps = 0;
        int u = A;
        for (int i = ceil(log2(n)); i >= 0; --i) {
            if(h[st0[u][i]] < h[C]){
                u = st0[u][i];
                jumps += (1 << i);
            }
        }
        // If the first sparse table is all we need, we're done
        if(st0[u][0] == C){
            return jumps + 1;
        } else {
            // Otherwise start at the next sparse table
        for (int i = ceil(log2(n)); i >= 0; --i) {
                if(h[st1[u][i]] < h[C]){
                    u = st1[u][i];
                    jumps += (1 << i);
                }
            }
    
            if(st1[u][0] == C){
                return jumps + 1;
            }
        }
        return -1;
    // Otherwise, if it's subtask 1, we do our dumb solution
    } else if(subtask1){
        return C - B;
    // Subtask 6 - wavelet tree
    } else if(C == D){
        cout << 100;
        query.clear();
        //rangeq(C, A, B, 0, n);
 
        for(int i = 0; i < query.size(); i ++){
            cout << query[i] << " ";
        } cout << "\n";
 
 
         // Now it's only subtask 2, 3 or 4 so we do BFS
    } else {
        //return u;
    if(jumps){
        return jumps + 1;
    } else {
        return 0;
    }
        queue<pair<int, int>> q;
        for(int i = A; i <= B; i ++){
            q.push({0, i});
        }
        int node = 0; int dist = 0;
        vector<int> vi(n, 0);
        while(!q.empty()){
            node = q.front().second;
            dist = q.front().first;
            //cout << node << "\n";
            q.pop();    x   
            if(node >= C && node <= D){
                return dist;
                //break;
            }
            if(!vi[node]){
                vi[node] = 1;
                for(int i = 0; i < adj[node].size(); i ++){
                    q.push({dist + 1, adj[node][i]});
                }
            }
        }
 
        return -1;
    }
*/

컴파일 시 표준 에러 (stderr) 메시지

jumps.cpp: In function 'void build(int, int, int)':
jumps.cpp:65:26: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   65 |         for(int x = 0; x < tree[j].size(); x ++){
      |                        ~~^~~~~~~~~~~~~~~~
jumps.cpp: In function 'void dfs0(int, int)':
jumps.cpp:135:26: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
  135 |         for(int i = 0; i < radj[u].size(); i ++){
      |                        ~~^~~~~~~~~~~~~~~~
jumps.cpp:137:30: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
  137 |             for(int j = 0; j < adj[radj[u][i]].size(); j ++){
      |                            ~~^~~~~~~~~~~~~~~~~~~~~~~~
jumps.cpp: In function 'void dfs1(int, int)':
jumps.cpp:155:26: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
  155 |         for(int i = 0; i < radj[u].size(); i ++){
      |                        ~~^~~~~~~~~~~~~~~~
jumps.cpp:157:30: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
  157 |             for(int j = 0; j < adj[radj[u][i]].size(); j ++){
      |                            ~~^~~~~~~~~~~~~~~~~~~~~~~~
jumps.cpp: In function 'void init(int, std::vector<int>)':
jumps.cpp:224:22: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
  224 |     for(int i = 0; i < H.size(); i ++){
      |                    ~~^~~~~~~~~~

• Subtask 1: 0 / 4
• Subtask 2: 0 / 8
• Subtask 3: 0 / 13
• Subtask 4: 0 / 12
• Subtask 5: 23 / 23
• Subtask 6: 0 / 21
• Subtask 7: 0 / 19