Openjudge-2192-Zipper

最新推荐文章于 2022-08-25 17:22:13 发布

原创最新推荐文章于 2022-08-25 17:22:13 发布 · 500 阅读

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CC 4.0 BY-SA版权

本文介绍了一个字符串问题：如何判断第三个字符串是否能由前两个字符串按原顺序交替组合而成。使用递归深度优先搜索（DFS）的方法解决此问题，并提供了一段AC代码作为实现案例。

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总时间限制:

1000ms

内存限制:

65536kB

描述

Given three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The first two strings can be mixed arbitrarily, but each must stay in its original order.

For example, consider forming "tcraete" from "cat" and "tree":

String A: cat
String B: tree
String C: tcraete

As you can see, we can form the third string by alternating characters from the two strings. As a second example, consider forming "catrtee" from "cat" and "tree":

String A: cat
String B: tree
String C: catrtee

Finally, notice that it is impossible to form "cttaree" from "cat" and "tree".

输入

The first line of input contains a single positive integer from 1 through 1000. It represents the number of data sets to follow. The processing for each data set is identical. The data sets appear on the following lines, one data set per line.

For each data set, the line of input consists of three strings, separated by a single space. All strings are composed of upper and lower case letters only. The length of the third string is always the sum of the lengths of the first two strings. The first two strings will have lengths between 1 and 200 characters, inclusive.

输出

For each data set, print:

Data set n: yes

if the third string can be formed from the first two, or

Data set n: no

if it cannot. Of course n should be replaced by the data set number. See the sample output below for an example.

样例输入

3
cat tree tcraete
cat tree catrtee
cat tree cttaree

样例输出

Data set 1: yes
Data set 2: yes
Data set 3: no

思路：对于c字符串的每一个字符要么和a字串中的当前位置对应要么和b字符串中的当前位置对应。否则说明a,b不能在各自字符相对顺序不变的情况下组成c，即结果输出no.

而对于c中的每一个元素都进行相同的比较过程，则可以用递归的方式写。

也就是通过这道题我对于dfs有了一些了解，高兴。

模板：

AC代码：

#include<iostream>
#include<string.h>
using namespace std;
char a[202],b[202],c[402]; //注意c[]的大小，wa了好几次。
int visit[202][202];
int flag,la,lb,lc;
void dfs(int x,int y,int z)
{
	if (z==lc)         //目标状态
	{
		flag=1;
		return ;
	}
	if (flag)
		return ;
	if (visit[x][y])   //如果改点不合法
		return ;
	visit[x][y]=1;
	if (a[x]==c[z])      如果该点合法但又不是目标状态
		dfs(x+1,y,z+1);   //递归调用
	if (b[y]==c[z])
		dfs(x,y+1,z+1);
}
int main()
{
	int n,k=0;
	cin>>n;
	while(n--)
	{

    	cin>>a>>b>>c;
		memset(visit,0,sizeof(visit));
		flag=0;
		la=strlen(a);
		lb=strlen(b);
		lc=strlen(c);
		dfs(0,0,0);                // a,b字符串都从开头开始判断
		if (flag&&la+lb==lc)      //如果是yes的话，a和b的长度一定等于c的长度
			cout<<"Data set "<<++k<<": "<<"yes"<<endl;
		else
			cout<<"Data set "<<++k<<": "<<"no"<<endl;
	}
	return 0;
}

希望有点用吧。