UVA 10887 Concatenation of Languages 字符串hash

Posted 2020-09-29 Meek

题目链接：https://vjudge.net/problem/UVA-10887

题意：

　　给你两个集合A,B，任意组合成新的集合C（去重）

　　问你最后C集合大小

题解：

　　暴力

　　组成的新串hash起来

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
typedef unsigned long long ULL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 2e3+10, M = 1e3+20,inf = 2e9;
ULL mod = 10000019ULL;
int n,m;
char a[N][202],b[N][202];
map<ULL , int > mp;
int ans;
int check(int i,int j) {
    ULL now = 0;
    ULL ps = 1;
    for(int ii = 0; ii < strlen(a[i]); ++ii) {
        now = now + ps * (a[i][ii]) ;
        ps *= mod;
    }
    for(int ii = 0; ii < strlen(b[j]); ++ii) {
        now = now + ps * (b[j][ii]) ;
         ps *= mod;
    }
    if(mp[now]) return 1;
    else {
        ans++;
        mp[now] = 1;
        return 0;
    }
}
int main() {
    int T,cas = 1;
    scanf("%d",&T);
    while(T--) {
       mp.clear();
        char ch[30];
        scanf("%d%d",&n,&m);
        getchar();
        for(int i = 1; i <= n; ++i)
           gets(a[i]);
        for(int i = 1; i <= m; ++i)
           gets(b[i]);
        ans = 0;
        for(int i = 1; i <= n; ++i) {
            for(int j = 1; j <= m; ++j) {
                check(i,j);
            }
        }
        printf("Case %d: %d\n",cas++,ans);
    }
    return 0;
}