Submission #971176

#	Time	Username	Problem	Language	Result	Execution time	Memory
971176		dubabuba	Rainforest Jumps (APIO21_jumps)	C++14	0 / 100	161 ms	45880 KiB

jumps

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

using namespace std;

template<typename T> void ono_min(T &MIN, T CMP) { if(MIN > CMP) MIN = CMP; }
template<typename T> void ono_max(T &MAX, T CMP) { if(MAX < CMP) MAX = CMP; }

typedef pair<int, int> pii;
#define ff first
#define ss second
#define MP make_pair

const int mxk = 32;
const int mxn = 2e5 + 10;
int L[mxn], R[mxn], n;
vector<int> adj[mxn];
int dis[mxn], h[mxn];

int mx[mxk][mxn];
int mn[mxk][mxn];

bool mx_valid(int k, int i) { return 1 <= mx[k][i] && mx[k][i] <= n; }
bool mn_valid(int k, int i) { return 1 <= mn[k][i] && mn[k][i] <= n; }

void init(int N, vector<int> H) {
	memset(L, -1, sizeof L);
	memset(R, -1, sizeof R);
	n = N;

	stack<int> st; st.push(0);
	for(int i = 1; i < N; i++) {
		while(st.size() && H[st.top()] <= H[i])
			st.pop();
		if(st.size())
			L[i] = st.top();
		st.push(i);
	}

	stack<int> en; en.push(0);
	for(int i = N - 1; i >= 0; i--) {
		while(en.size() && H[en.top()] <= H[i])
			en.pop();
		if(en.size())
			R[i] = en.top();
		en.push(i);
	}

	for(int i = 0; i < N; i++) {
		h[i] = H[i];
		if(L[i] > -1) {
			adj[i].push_back(L[i]);
			if(mn[0][H[i]] == 0)
				mn[0][H[i]] = H[L[i]];
			ono_min(mn[0][H[i]], H[L[i]]);
			ono_max(mx[0][H[i]], H[L[i]]);
		}
		if(R[i] > -1) {
			adj[i].push_back(R[i]);
			if(mn[0][H[i]] == 0)
				mn[0][H[i]] = H[R[i]];
			ono_min(mn[0][H[i]], H[R[i]]);
			ono_max(mx[0][H[i]], H[R[i]]);
		}
	}

	for(int k = 1; k < mxk; k++)
	for(int i = 1; i <= n; i++) {
		int mnj = mn[k - 1][i];
		int mxj = mx[k - 1][i];

		if(mnj) mn[k][i] = mn[k - 1][mnj];
		if(mxj) mx[k][i] = mx[k - 1][mxj];
	}
}

int minimum_jumps(int A, int B, int C, int D) {
	if(A != B) exit(1);
	if(C != D) exit(1);
	if(h[A] > h[C]) return -1;

	int s = h[A], f = h[C], ans = 0;
	for(int k = mxk - 1; k >= 0; k--)
	if(mx[k][s] && mx[k][s] <= f) {
		s = mx[k][s];
		ans ^= (1 << k);
	}

	for(int k = mxk - 1; k >= 0; k--)
	if(mn[k][s] && mn[k][s] <= f) {
		s = mn[k][s];
		ans ^= (1 << k);
	}

	if(f == s) return ans;
	return -1;
}

• 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