POJ 1088 解题报告

按照http://blog.csdn.net/thestoryofsnow/article/details/40942009的顺序刷，果然轻松愉快。这道题就是DFS，同时记录结果避免重复。做完后看了看解法，大致如此，大同小异，表示对poj上0ms过的很惊奇，不知如何做到的。

1088

Accepted

336K

63MS

C++

1683B

/* 
ID: thestor1 
LANG: C++ 
TASK: poj1088 
*/
#include <iostream>
#include <fstream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <climits>
#include <cassert>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <cassert>

using namespace std;

int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};

void DFS(int r, int c, const int R, const int C, std::vector<std::vector<int> > &heights, std::vector<std::vector<int> > &seqs)
{
	seqs[r][c] = 1;

	int nr, nc;
	for (int d = 0; d < 4; ++d)
	{
		nr = r + dx[d], nc = c + dy[d];
		if (0 <= nr && nr < R && 0 <= nc && nc < C && heights[r][c] > heights[nr][nc])
		{
			if (seqs[nr][nc] < 0)
			{
				DFS(nr, nc, R, C, heights, seqs);
			}
			if (seqs[nr][nc] + 1 > seqs[r][c])
			{
				seqs[r][c] = seqs[nr][nc] + 1;
			}
		}
	}
}

int main()
{
	int R, C;
	cin >> R >> C;
	std::vector<std::vector<int> > heights(R, std::vector<int>(C, 0));
	std::vector<std::vector<int> > seqs(R, std::vector<int>(C, -1));
	for (int r = 0; r < R; ++r)
	{
		for (int c = 0; c < C; ++c)
		{
			cin >> heights[r][c];
		}
	}

	// cout << "heights:" << endl;
	// for (int r = 0; r < R; ++r)
	// {
	// 	for (int c = 0; c < C; ++c)
	// 	{
	// 		cout << heights[r][c];
	// 	}
	// 	cout << endl;
	// }

	int maxseq = 1;
	for (int r = 0; r < R; ++r)
	{
		for (int c = 0; c < C; ++c)
		{
			if (seqs[r][c] < 0)
			{
				DFS(r, c, R, C, heights, seqs);
			}
			if (seqs[r][c] > maxseq)
			{
				maxseq = seqs[r][c];
			}
		}
	}

	cout << maxseq << endl;

	return 0;  
}