uva 10340 All in All(子串）

All in All
Input: standard input
Output: standard output
Time Limit: 2 seconds
Memory Limit: 32 MB
You have devised a new encryption technique which encodes a message by inserting between its characters randomly generated strings in a clever way. Because of pending patent issues we will not discuss in detail how the strings are generated and inserted
into the original message. To validate your method, however, it is necessary to write a program that checks if the message is really encoded in the final string.

Given two strings s and t, you have to decide whether s is a subsequence of t, i.e. if you can remove characters from t such that the concatenation of the remaining characters is s.

Input Specification

The input contains several testcases. Each is specified by two strings s, t of alphanumeric ASCII characters separated by whitespace. Input is terminated by EOF.

Output Specification

For each test case output, if s is a subsequence of t.

Sample Input

sequence subsequence

person compression

VERDI vivaVittorioEmanueleReDiItalia

caseDoesMatter CaseDoesMatter

Sample Output

Yes

No

Yes

No

题目大意：给出两个字符串，判断串1是否为串2的子串，这里的子串不要求在串2中连续，但是要保证顺序一致。
解题思路：递归遍历。

#include <stdio.h>
#include <string.h>

const int N = 1000005;
char f
, s
;
int n, m, vis
;

bool solve(int a, int b) {
if (b >= m)	return true;

for (int i = a; i < n; i++) {
if (f[i] == s[b]) {
if (solve(i + 1, b + 1))
return true;
}
}
return false;
}

int main() {
while (scanf("%s%s", s, f) == 2) {
n = strlen(f), m = strlen(s);
printf("%s\n", solve(0, 0) ? "Yes" : "No");
}
return 0;
}