CodeForces 508A Pasha and Pixels

 Pasha and Pixels
Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

Pasha loves his phone and also putting his hair up... But the hair is now irrelevant.

Pasha has installed a new game to his phone. The goal of the game is following. There is a rectangular field consisting of n row with mpixels in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black. Pasha loses the game when a 2 × 2 square consisting of black pixels is formed.

Pasha has made a plan of k moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbers i and j, denoting respectively the row and the column of the pixel to be colored on the current move.

Determine whether Pasha loses if he acts in accordance with his plan, and if he does, on what move the 2 × 2 square consisting of black pixels is formed.

Input

The first line of the input contains three integers n, m, k (1 ≤ n, m ≤ 10001 ≤ k ≤ 105) — the number of rows, the number of columns and the number of moves that Pasha is going to perform.

The next k lines contain Pasha's moves in the order he makes them. Each line contains two integers i and j (1 ≤ i ≤ n1 ≤ j ≤ m), representing the row number and column number of the pixel that was painted during a move.

Output

If Pasha loses, print the number of the move when the 2 × 2 square consisting of black pixels is formed.

If Pasha doesn't lose, that is, no 2 × 2 square consisting of black pixels is formed during the given k moves, print 0.

Sample Input

Input
2 2 4
1 1
1 2
2 1
2 2
Output
4
Input
2 3 6
2 3
2 2
1 3
2 2
1 2
1 1
Output
5
Input
5 3 7
2 3
1 2
1 1
4 1
3 1
5 3
3 2
Output
0



题意:一个矩阵,刚开始输入它的行和列, 都是白色,一个数k, 来涂色, 接下来是k个输入,每次一个i, j;  代表位置, 将i, j 的位置涂黑, 涂过黑色就不变了,只要出现2×2 被涂黑过,就输出到现在的步数;当然k个输入都要输入完毕才可以,如果k个输入都输入完了,也没出现2 *2 的矩阵是黑色,就输出0;


#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <stack>
using namespace std;
bool num[1010][1010];
int n, m, k;

bool exist(int x, int y){
	if(x - 1 > 0 && y - 1 > 0 && num[x -1][y -1] && num[x - 1][y] && num[x][y - 1])	return true;
	else if(x -1 > 0 && y + 1 <= m && num[x -1][y] && num[x - 1][y + 1] && num[x][y + 1]) return true;
	else if(x + 1 <= n && y - 1 > 0 && num[x][y - 1] && num[x + 1][y - 1] & num[x + 1][y])	return true;
	else if(x + 1 <= n && y + 1 <= m && num[x][y + 1] && num[x + 1][y + 1] && num[x + 1][y])	return true;
	else 
		return false;
}
int main(){
	
	while(~scanf("%d%d%d", &n, &m, &k)){
	
		memset(num, false, sizeof(num));
		int i, j;
		int step = 0;
		int flag = 0;
		while(k--){
			cin >> i >> j;
			num[i][j] = true;
			if(flag == 0){
				++step;
				if(exist(i, j)){
					flag = 1;
				}
			}
		}
		if(flag == 0){
			cout << "0" << endl;
		}
		else 
			cout << step << endl;
	}
	return 0;
}






















内容概要:本文介绍了一个基于MATLAB实现的无人机三维路径规划项目,采用蚁群算法(ACO)与多层感知机(MLP)相结合的混合模型(ACO-MLP)。该模型通过三维环境离散化建模,利用ACO进行全局路径搜索,并引入MLP对环境特征进行自适应学习与启发因子优化,实现路径的动态调整与多目标优化。项目解决了高维空间建模、动态障碍规避、局部最优陷阱、算法实时性及多目标权衡等关键技术难题,结合并行计算与参数自适应机制,提升了路径规划的智能性、安全性和工程适用性。文中提供了详细的模型架构、核心算法流程及MATLAB代码示例,涵盖空间建模、信息素更新、MLP训练与融合优化等关键步骤。; 适合人群:具备一定MATLAB编程基础,熟悉智能优化算法与神经网络的高校学生、科研人员及从事无人机路径规划相关工作的工程师;适合从事智能无人系统、自动驾驶、机器人导航等领域的研究人员; 使用场景及目标:①应用于复杂三维环境下的无人机路径规划,如城市物流、灾害救援、军事侦察等场景;②实现飞行安全、能耗优化、路径平滑与实时避障等多目标协同优化;③为智能无人系统的自主决策与环境适应能力提供算法支持; 阅读建议:此资源结合理论模型与MATLAB实践,建议读者在理解ACO与MLP基本原理的基础上,结合代码示例进行仿真调试,重点关注ACO-MLP融合机制、多目标优化函数设计及参数自适应策略的实现,以深入掌握混合智能算法在工程中的应用方法。
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