UVA - 12338 Anti-Rhyme Pairs 二分+hash

Posted 2020-09-29 Meek

题目链接：https://vjudge.net/problem/UVA-12338

题意：

　　给你n个串

　　问你任意两个串的最大公共前缀长度是多少

题解：

　　二分+hash

　　思路很明显，我最近用来写hash

　　很鸡肋

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>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 = 2e5+10, M = 1e3+20,inf = 2e9;

ULL mod = 10000019ULL;
vector<ULL > has[N];
int n,q;
char ch[N];
int main() {
    int T,cas = 1;
    scanf("%d",&T);
    while(T--) {
        scanf("%d",&n);
        for(int i = 1; i <= n; ++i) has[i].clear();
        for(int i = 1; i <= n; ++i) {
            scanf("%s",ch);
            int len = strlen(ch);
            ULL now = 0;
            ULL ps = 1;
            for(int j = 0; j < len; ++j) {
                has[i].push_back(0);
                has[i][j] = now + ps * ch[j];
                now = has[i][j];
                ps *= mod;
            }
        }
        scanf("%d",&q);
        printf("Case %d:\n",cas++);
        for(int i = 1; i <= q; ++i) {
            int x,y;
            scanf("%d%d",&x,&y);
            int ans = -1;
            int ll = 0, rr = min(has[x].size(),has[y].size()) - 1;
            while(ll <= rr) {
                if(has[x][mid] == has[y][mid])
                    ans = mid,ll = mid+1;
                else rr = mid - 1;
            }
            printf("%d\n",ans+1);
        }
    }
    return 0;
}