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Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12387 Accepted Submission(s): 3022
无意中看到这题，就切了一下，感觉这题很适合刚刚接触散列表和字符串处理的朋友，直接对字符串hash就可以了，用map会MlE。
AC代码：
#include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> using namespace std; const int MAX=210000; const int mod=100007; struct Node{ Node* ne; char st[81]; }haxi[MAX],*h[mod],*cur; unsigned int BKDHash(char* s){ unsigned int seed=131; unsigned int ret=0; while(*s) ret=ret*seed+*s++; return (ret&0x7FFFFFFF)%mod; } int getId(char* s){ int code=BKDHash(s); Node* ptr=h[code]; while(ptr){ if(strcmp(ptr->st,s)==0) return ptr-haxi; else ptr=ptr->ne; } strcpy(cur->st,s); cur->ne=h[code]; h[code]=cur++; return cur-haxi-1; } int find(char* s){ int code=BKDHash(s); Node* ptr=h[code]; while(ptr){ if(strcmp(ptr->st,s)==0){ return ptr-haxi; }else{ ptr=ptr->ne; } } return -1; } int main(){ char s[100],*p; int id,n; cur=haxi; memset(h,0,sizeof(h)); while(scanf("%s",s),s[0]!=‘@‘){ getId(s); getchar(); gets(s); getId(s); } scanf("%d",&n); gets(s); while(n--){ gets(s); id=find(s); if(id==-1){ puts("what?"); }else{ p=haxi[id^1].st; if(p[0]!=‘[‘) puts(p); else{ p++; while(*p!=‘]‘){ putchar(*p++); } puts(""); } } } return 0; }
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原文地址：http://www.cnblogs.com/shenben/p/5751085.html
