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

 /*

     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 ;

 }