Submission #336786

#TimeUsernameProblemLanguageResultExecution timeMemory
336786super_j6Abduction 2 (JOI17_abduction2)C++14
27 / 100
50 ms11884 KiB
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <array>
#include <unordered_map>
using namespace std;
#define endl '\n'
#define ll long long
#define pi pair<int, int>
#define f first
#define s second
 
typedef array<int, 2> T;
 
T operator+(T x, T y){
	return {x[0] + y[0], x[1] + y[1]};
}
 
const int mxn = 50000, k = 4;
const T d[k] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
int n[2], q;
int a[2][mxn];
unordered_map<int, int> dp[k][mxn];

bool f(T x){
	for(int i = 0; i < 2; i++) if(x[i] < 0 || x[i] >= n[i]) return 0;
	return 1;
}
 
int sol(T x, int y){
	int ret = 0, s = 0;
	for(; f(x) && a[!(y & 1)][x[!(y & 1)]] > a[y & 1][x[y & 1]]; x = x + d[y]) s++;
	if(!f(x)) return s;
	if(dp[y][x[0]].count(x[1])) return dp[y][x[0]][x[1]] + s;
	for(int i = 0; i < 2; i++){
		int z = (y + 2 * i + 1) % k;
		ret = max(ret, sol(x + d[z], z));
	}
	dp[y][x[0]][x[1]] = ++ret;
	return ret + s;
}
 
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n[0] >> n[1] >> q;
	
	if(max(n[0], n[1]) == mxn) return 0;
	
	for(int i = 0; i < 2; i++)
	for(int j = 0; j < n[i]; j++){
		cin >> a[i][j];
	}
	
	while(q--){
		T x;
		for(int i = 0; i < 2; i++) cin >> x[i], x[i]--;
		int ret = 0;
		for(int i = 0; i < k; i++) ret = max(ret, sol(x + d[i], i));
		cout << ret << endl;
	}
 
	return 0;
}
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