rom through esp 代碼 bsp see itl lan integer

原題鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=1501

一：原題內容

Problem 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

二：分析理解

第三個字符串能否由前兩個字符串依照原有順序不變的原則交叉構成。

須要註意的是，visit數組元素值為1時。表示該位置已被訪問過，下次無需訪問。

三：AC代碼

#define _CRT_SECURE_NO_DEPRECATE 
#define _CRT_SECURE_CPP_OVERLOAD_STANDARD_NAMES 1  

#include<iostream>
#include<string>
#include<string.h>
using namespace std;

string str1, str2, str3;
int len1, len2, len3;
bool flag;//為真時，表示能夠輸出“yes”

int visit[201][201];//標記數組。默認都是0

void DFS(int i, int j, int k);

int main()
{
	int N;
	cin >> N;
	for (int i = 1; i <= N; i++)
	{
		memset(visit, 0, sizeof(visit));
		flag = false;
		cin >> str1 >> str2 >> str3;

		len1 = str1.length();
		len2 = str2.length();
		len3 = str3.length();


		DFS(0, 0, 0);

		if (flag)
			cout << "Data set " << i << ": " << "yes\n";
		else
			cout << "Data set " << i << ": " << "no\n";
	}

	return 0;
}

void DFS(int i, int j, int k)
{
	if (flag || visit[i][j])//假設為真或該點已被訪問過
		return;

	if (k == len3)//由於依據題意len1+len2=len3
	{
		flag = true;
		return;
	}

	visit[i][j] = 1;

	if (i < len1 && str1[i] == str3[k])
		DFS(i + 1, j, k + 1);
	if (j < len2 && str2[j] == str3[k])
		DFS(i, j + 1, k + 1);

}

hdu1501 Zipper--DFS