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

题目传送门

 /*
     题意：给出一系列名字变化，问最后初始的名字变成了什么
     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
                 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:)
 */
 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <string>
 #include <map>
 #include <ctime>
 using namespace std;

 const int MAXN = 1e3 + ;
 const int INF = 0x3f3f3f3f;
 string old[MAXN];
 string change[MAXN];
 string ans[MAXN];
 int pos[MAXN];

 int main(void)        //Codeforces Round #285 (Div. 2) B. Misha and Changing Handles
 {
     freopen ("B.in", "r", stdin);

     int n;
     while (scanf ("%d", &n) == )
     {
         memset (pos, , sizeof (pos));
         string tmp;    int m = ;
         for (int i=; i<=n; ++i)
         {
             cin >> tmp; cin >> change[i];
             bool ok = false;
             for (int j=; j<i; ++j)
             {
                 if (tmp == change[j])
                 {
                     ok = true;
                     ans[pos[j]] = change[i];
                     pos[i] = pos[j];    break;
                 }
             }
             if (!ok)
             {
                 old[++m] = tmp; ans[m] = change[i];    pos[i] = m;
             }
         }

         printf ("%d\n", m);
         for (int i=; i<=m; ++i)
         {
             cout << old[i] << " " << ans[i] << endl;
         }
     }

     // cout << "time: " << (double) clock () / CLOCKS_PER_SEC * 1000 << " ms\n";

     return ;
 }