hdu1501&&poj2192 Zipper(DFS)

时间：2017-04-23 16:58:22 阅读：151 评论：0 收藏：0 [点我收藏+]

标签：ase while pid case fine == int cst search

转载请注明出处：http://blog.csdn.net/u012860063

题目链接

HDU：http://acm.hdu.edu.cn/showproblem.php?pid=1501

Zipper

Description

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".

Input

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.

Output

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.

Sample Input

3
cat tree tcraete
cat tree catrtee
cat tree cttaree

Sample Output

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

Source

Pacific Northwest 2004

题意：就是找在不变换字符串1和字符串2的字母的顺序的情况下，是否能组成字符串3。

代码例如以下：

#include <cstdio>
#include <cstring> 
#define N 247
char a[N], b[N], c[N+N];
int len1, len2, len3;
int dfs(int x, int y, int sum)
{
	if(sum >len3)
	return 1;
	if(a[x]!=c[sum] && b[y]!=c[sum])
	return 0;
	if(a[x]==c[sum] && dfs(x+1,y,sum+1))
	return 1;
	if(b[y]==c[sum] && dfs(x,y+1,sum+1))
	return 1;
	return 0;
}
int main()
{
	int t, cas = 0;
	while(~scanf("%d",&t))
	{
		while(t--)
		{
			scanf("%s%s%s",a+1,b+1,c+1);//从字符串下标为1開始输入 
			len1 = strlen(a+1);
			len2 = strlen(b+1);
			len3 = strlen(c+1);
			int flag = 0;
			if(c[len3]==a[len1] || c[len3]==b[len2])//优化，假设C最后的字母是a的最后一个 
			{										//或者b的最后一个才有可能是由它们组成的 
				flag = dfs(1, 1, 1);
			}
			if(flag == 1)
			printf("Data set %d: yes\n",++cas);
			else
			printf("Data set %d: no\n",++cas);
		}
	}
	return 0;
}

hdu1501&&poj2192 Zipper(DFS)

标签：ase while pid case fine == int cst search

原文地址：http://www.cnblogs.com/wgwyanfs/p/6752745.html

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