字符串处理 Codeforces Round #285 (Div. 2) B. Misha and Changing Handles

题目传送门

 1 /*
 2     题意：给出一系列名字变化，问最后初始的名字变成了什么
 3     字符串处理：每一次输入到之前的找相印的名字，若没有，则是初始的，pos[m] 数组记录初始位置
 4                 在每一次更新时都把初始pos加上去，那么就保证更新了初始的名字，这也是唯一要思考的地方了:)
 5 */
 6 #include <cstdio>
 7 #include <iostream>
 8 #include <cstring>
 9 #include <algorithm>
10 #include <cmath>
11 #include <string>
12 #include <map>
13 #include <ctime>
14 using namespace std;
15
16 const int MAXN = 1e3 + 10;
17 const int INF = 0x3f3f3f3f;
18 string old[MAXN];
19 string change[MAXN];
20 string ans[MAXN];
21 int pos[MAXN];
22
23 int main(void)        //Codeforces Round #285 (Div. 2) B. Misha and Changing Handles
24 {
25     freopen ("B.in", "r", stdin);
26
27     int n;
28     while (scanf ("%d", &n) == 1)
29     {
30         memset (pos, 0, sizeof (pos));
31         string tmp;    int m = 0;
32         for (int i=1; i<=n; ++i)
33         {
34             cin >> tmp; cin >> change[i];
35             bool ok = false;
36             for (int j=1; j<i; ++j)
37             {
38                 if (tmp == change[j])
39                 {
40                     ok = true;
41                     ans[pos[j]] = change[i];
42                     pos[i] = pos[j];    break;
43                 }
44             }
45             if (!ok)
46             {
47                 old[++m] = tmp; ans[m] = change[i];    pos[i] = m;
48             }
49         }
50
51         printf ("%d\n", m);
52         for (int i=1; i<=m; ++i)
53         {
54             cout << old[i] << " " << ans[i] << endl;
55         }
56     }
57
58     // cout << "time: " << (double) clock () / CLOCKS_PER_SEC * 1000 << " ms\n";
59
60     return 0;
61 }