ZOJ 2100 Seeding

最新推荐文章于 2020-02-17 08:35:13 发布

Calm微笑

最新推荐文章于 2020-02-17 08:35:13 发布

阅读量419

点赞数 1

CC 4.0 BY-SA版权

分类专栏： DFS

本文链接：https://blog.youkuaiyun.com/yao1373446012/article/details/50855573

DFS 专栏收录该内容

30 篇文章

订阅专栏

本文深入探讨了大数据开发领域的关键技术，包括Hadoop、Spark、Flink等，并结合实际案例阐述了这些技术在数据处理、分析和存储方面的应用。通过详细解析数据流处理流程和系统架构，为读者提供了一套全面的大数据开发解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Seeding

Time Limit: 2 Seconds Memory Limit: 65536 KB

It is spring time and farmers have to plant seeds in the field. Tom has a nice field, which is a rectangle with n * m squares. There are big stones in some of the squares.

Tom has a seeding-machine. At the beginning, the machine lies in the top left corner of the field. After the machine finishes one square, Tom drives it into an adjacent square, and continues seeding. In order to protect the machine, Tom will not drive it into a square that contains stones. It is not allowed to drive the machine into a square that been seeded before, either.

Tom wants to seed all the squares that do not contain stones. Is it possible?

Input

The first line of each test case contains two integers n and m that denote the size of the field. (1 < n, m < 7) The next n lines give the field, each of which contains m characters. 'S' is a square with stones, and '.' is a square without stones.

Input is terminated with two 0's. This case is not to be processed.

Output

For each test case, print "YES" if Tom can make it, or "NO" otherwise.

Sample Input

4 4
.S..
.S..
....
....
4 4
....
...S
....
...S
0 0

Sample Output

YES
NO

Author: ZHANG, Zheng

Source: Zhejiang University Local Contest 2004

这道题刚开始写好后，运行一下没什么问题，但是提交之后就是WA，找了半天也没找到错误，于是就把代码删了重新写了一遍，然后就AC了。。。。

这道题主要的地方就是一个回溯问题，即如果此路不通就把石头重新变回土地，最开始学时懵懵懂懂，并且也没有深入去理解，不过现在理解起来容易多了

题意：石头"S"不能过，用过的土地"."不能再用，能否把土地一次性种上种子

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int i,j,m,n,a,flag,k;
char map[10][10];
void dfs(int x,int y)
{
	if(k==n*m)
	{
		flag=1;
		return ;
	}
	if(x<0||x>=m||y<0||y>=n||map[x][y]=='S')
	return ;
	if(flag)
	return ;
	a++;
	if(a>1500)//这是个很值得学习的地方，如果查找的次数，超过了这个，就不用再继续下去了，节省了代码运行时间 
	return ;
	map[x][y]='S';
	k++;
	dfs(x+1,y);
	dfs(x-1,y);
	dfs(x,y+1);
	dfs(x,y-1);
	k--;
	map[x][y]='.';
}
int main()
{
	while(scanf("%d%d",&m,&n),m+n)
	{
		a=0;
		flag=0;
		k=0;
		for(i=0;i<m;i++)
		scanf("%s",map[i]);
		for(i=0;i<m;i++)
		for(j=0;j<n;j++)
		if(map[i][j]=='S')
		k++;
		dfs(0,0);
		if(flag)
		printf("YES\n");
		else
		printf("NO\n"); 
	}
	return 0;
}