Submission #718695

#	Time	Username	Problem	Language	Result	Execution time	Memory
718695		lam	Rainforest Jumps (APIO21_jumps)	C++14	0 / 100	277 ms	44040 KiB

jumps

#include "jumps.h"
#include <bits/stdc++.h>
#include <vector>
using namespace std;
const int maxn = 2e5 + 10;
int n;
int a[maxn],b[maxn],d[maxn];
vector <int> adj[maxn];
int par[20][maxn];
int f[20][maxn],g[20][maxn];
void create()
{
    for (int i=0; i<n; i++) f[0][i] = b[i];
    for (int j=1; (1<<j)<=n; j++)
        for (int i=0; i+(1<<j)-1<n; i++)
            f[j][i] = max(f[j-1][i],f[j-1][i+(1<<(j-1))]);
}

int get(int l, int r)
{
    if (l>r) return -1;
    int k=log2(r-l+1);
    return max(f[k][l],f[k][r-(1<<k)+1]);
}

void addedge(int u, int v)
{
    adj[u].push_back(v);
}
void init(int N, std::vector<int> H) {
    n=N;
    for (int i=0; i<n; i++) a[i]=H[i],adj[i].clear();
    deque<int> dq;
    for (int i=n-1; i>=0; i--)
    {
        while (!dq.empty()&&a[dq.back()] < a[i]) dq.pop_back();
        if (!dq.empty()) b[i] = dq.back(), d[i]=d[dq.back()]+1;
        else b[i] = -1,d[i]=0;
        dq.push_back(i);
    }
    while (!dq.empty()) dq.pop_back();
    for (int i=0; i<n; i++)
    {
        while (!dq.empty()&&a[dq.back()] < a[i]) dq.pop_back();
        if (!dq.empty()) par[0][i] = dq.back(), g[0][i] = min(d[i],d[dq.back()]+1);
        else par[0][i]=-1, g[0][i] = d[i];
        dq.push_back(i);
    }
    create();
    for (int i=0; i<n; i++)
        for (int j=1; j<20; j++)
            if (par[j-1][i]==-1) par[j][i]=-1,g[j][i] = d[i];
            else
            {
                par[j][i]=par[j-1][par[j-1][i]];
                g[j][i] = min(g[j-1][i],(1<<(j-1)) + g[j-1][par[j-1][i]]);
            }
}

int minimum_jumps(int A, int B, int C, int D) {
    if (A==C) return 0;
    if (a[A] > a[C]) return -1;
    int smx = get(B+1,C-1);
    if (smx > a[C]) return -1;
    int ans = 1e9;
    ans = d[A] - d[C];
    int u=A;
    int num = 0;
    for (int i=19; i>=0; i--)
        if (par[i][u]!=-1&&a[par[i][u]] < a[C])
        {
            ans = min(ans,g[i][u]-d[C]+num);
            num+=(1<<i);
            u=par[i][u];
        }
    return ans;

}

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