UESTC_韩爷的梦 2015 UESTC Training for Search Algorithm & String<Problem N>
2015-05-13 00:59
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N-韩爷的梦
TimeLimit:200/100MS(Java/Others)MemoryLimit:1300/1300KB(Java/Others)SubmitStatus
一天,韩爷去百度面试,面试官给了他这么一个问题。
给你2万个字符串,每个字符串长度都是100,然后把2万个字符串丢入一个set<string>g中,问最终set里含有多少个元素?
g是一个用来存储字符串、具有去重功能的容器,即相同字符串在g中只能保留一个。
两个字符串相等,当且仅当,长度一样且对应位置的字符都一样。
韩爷前晚没睡好,随手写了一个程序交给面试官,然后就gg了。
#include<iostream> #include<string> #include<set> usingnamespacestd; strings; set<string>g; intmain(){ for(intk=1;k<=20000;k++){ cin>>s; g.insert(s); } cout<<g.size()<<endl; return0; }
韩爷醒来之后,发现这只是一个梦(还好只是个梦)。他回忆起梦中的面试官给他的内存限制和时间限制非常低,这么做肯定过不了,那么,现在你不在梦中,你能解决这个问题么?
Input
单case每个case有且只有2万行,每一行包含一个字符串,每行字符串的长度都为100
(样例除外)
字符集:大写英文字母(A-Z),小写英文字母(a-z),数字(0-9)
Output
输出一个整数,表示最终set里含有多少个元素。Sampleinputandoutput
SampleInput | SampleOutput |
---|---|
aaAa aaAa bbbb 1234 bbbb bbbb ee09 | 4 |
Hint
样例只是样例,不在test中
注意时间限制和内存限制非常低
解题报告:
直接上哈希即可,双哈希单哈希都可以过,如果过不了,请换素数。。。。囧
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<cmath> usingnamespacestd; constintmaxn=2e4+100; charstr[maxn]; voidhashinit(int&x1,int&x2) { x1=0x7FED7FED,x2=1; intp1=1526597; intp2=89834777; intmod1=1e9+7; intmod2=1e9+9; unsignedintx3=0x23322322; for(inti=1;i<=100;++i) { intval=str[i]; x1=(x1*p1+val)%mod1; } for(inti=1;i<=100;++i) { intval=str[i]; x2=(x2*p2+val)%mod2; } } intmain(intargc,char*argv[]) { inthash1[maxn]; inthash2[maxn]; intsize=0; for(inti=1;i<=20000;++i) { scanf("%s",str+1); intq1,q2; hashinit(q1,q2); hash1[size]=q1, hash2[size++]=q2; } sort(hash1,hash1+size); sort(hash2,hash2+size); intc1=unique(hash1,hash1+size)-hash1; intc2=unique(hash2,hash2+size)-hash2; printf("%d\n",min(c1,c2)); return0; }
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