标签：
Time Limit: 20 Sec Memory Limit: 256 MB
http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=114147
The input consists of:
one line with one integer n (1≤n≤105), the number of submissions;
n lines, each with a result of the judging by DOMjudge, in arbitrary order;
n lines, each with a result of the judging by Kattis, in arbitrary order.
Each result is a string of length between 5 and 15 characters (inclusive) consisting of lowercase letters.
Output one line with the maximum number of judging results that could have been the same for both systems.
5
correct
wronganswer
correct
correct
timelimit
wronganswer
correct
timelimit
correct
timelimit
4
题意
有两个机器，每个机器都会返回n个串，然后问你有多少个串是在两个地方都出现过
题解：
双hash一下，然后用map存一下，然后搞一搞就好了……
代码：
//qscqesze #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define maxn 200001 #define mod 10007 #define eps 1e-9 int Num; char CH[20]; //const int inf=0x7fffffff; //нчоч╢С const int inf=0x3f3f3f3f; /* inline void P(int x) { Num=0;if(!x){putchar(‘0‘);puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts(""); } */ inline ll read() { int x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } inline void P(int x) { Num=0;if(!x){putchar(‘0‘);puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts(""); } //************************************************************************************** int get_hash(char *key) { ll N=1998585857; long long h=0; while(*key) h=(h*127+(*key++)+N)%N; return h%N; } int get_hash2(char *key) { ll N=127398127; long long h=0; while(*key) h=(h*127+(*key++)+N)%N; return h%N; } char s[20]; map< pair<int,int> ,int>H; int main() { int n=read(); for(int i=0;i<n;i++) { scanf("%s",s); pair<int,int> a; a.first=get_hash(s); a.second=get_hash2(s); H[a]++; } int ans=0; for(int i=0;i<n;i++) { scanf("%s",s); pair<int,int> a; a.first=get_hash(s); a.second=get_hash2(s); if(H[a]) { ans++; H[a]--; } } printf("%d\n",ans); }
标签：
原文地址：http://www.cnblogs.com/qscqesze/p/4510344.html
