String processing codeforces Round #285 (Div. 2) b. Misha and changing Handles

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1 /*2 Test Instructions: Give a series of name changes, ask what the last name became3 string Processing: Each time the input to the previous name of the printing, if not, it is the initial, Pos[m] array record the initial position4 the initial POS is added to each update, so the initial name is guaranteed to be updated, which is the only place to think about:)5 */6#include <cstdio>7#include <iostream>8#include <cstring>9#include <algorithm>Ten#include <cmath> One#include <string> A#include <map> -#include <ctime> - using namespacestd; the  - Const intMAXN = 1e3 +Ten; - Const intINF =0x3f3f3f3f; - stringOLD[MAXN]; + stringCHANGE[MAXN]; - stringANS[MAXN]; + intPOS[MAXN]; A  at intMainvoid)//codeforces Round #285 (Div. 2) b. Misha and changing Handles - { -Freopen ("b.in","R", stdin); -  -     intN; -      while(SCANF ("%d", &n) = =1) in     { -memset (POS,0,sizeof(POS)); to         stringtmpintm =0; +          for(intI=1; i<=n; ++i) -         { theCIN >> TMP; CIN >>Change[i]; *             BOOLOK =false; $              for(intj=1; j<i; ++j)Panax Notoginseng             { -                 if(TMP = =Change[j]) the                 { +OK =true; AANS[POS[J]] =Change[i]; thePos[i] = Pos[j]; Break; +                 } -             } $             if(!OK) $             { -OLD[++M] = tmp;    ANS[M] = Change[i]; Pos[i] =m; -             } the         } - Wuyiprintf ("%d\n", m); the          for(intI=1; i<=m; ++i) -         { Wucout << Old[i] <<" "<< Ans[i] <<Endl; -         } About     } $  -     //cout << "Time:" << (double) clock ()/clocks_per_sec * + << "ms\n"; -  -     return 0; A}

String processing codeforces Round #285 (Div. 2) b. Misha and changing Handles