Hopscotch POJ 3050

标签：dfs

1.题目描述：点击打开链接

2.解题思路：本题利用DFS解决。对每一个格子都进行dfs，当cur==6时，将合成的整数放入set中，最后输出set的大小即可。

3.代码：

#define _CRT_SECURE_NO_WARNINGS 
#include<iostream>
#include<algorithm>
#include<string>
#include<sstream>
#include<set>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<deque>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<functional>
using namespace std;

#define N 6
int g[N][N];
int A[N];
int dx[] = { -1, 1, 0, 0 };
int dy[] = { 0, 0, -1, 1 };
set<int>base;
bool inside(int x, int y){ return (x >= 0 && y >= 0 && x < 5 && y < 5); }
void dfs(int x, int y, int cur)
{

	if (cur == 6)
	{
		int sum = 0;
		for (int i = 0; i < cur; i++)
			sum = sum * 10 + A[i];
		base.insert(sum);
	}
	else
	{
		A[cur] = g[x][y];
		for (int i = 0; i < 4; i++)
		{
			int nx = x + dx[i], ny = y + dy[i];
			if (inside(nx, ny))
				dfs(nx, ny, cur + 1);
		}
	}
}
int main()
{
	//freopen("t.txt", "r", stdin);
	string str;
	while (getline(cin,str))
	{
		base.clear();
		memset(A, 0, sizeof(A));
		stringstream ss(str);
		int x, i = 0;
		while (ss >> x)g[0][i++] = x;
		for (int i = 1; i < 5; i++)
		{
			getline(cin, str);
			int j = 0;
			stringstream ss(str);
			while (ss >> x)g[i][j++] = x;
		}
		for (int i = 0; i < 5;i++)
		for (int j = 0; j < 5; j++)
			dfs(i, j, 0);
		cout << base.size() << endl;
	}
	return 0;
}

Hopscotch POJ 3050

标签：dfs

原文地址：http://blog.csdn.net/u014800748/article/details/44941741