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 
#include 

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

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;
}