POJ-2774 字符串哈希+二分

题意：给出s，t两个字符串，求最长公共子串的长度

思路：首先二分答案x，预处理出s中长度为x的子串哈希值并排序，在t中枚举起点，二分查找t[i...i+x]的哈希值

二分查找直接用binary_search（）函数

复杂度其实是nlog方

 

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define LL long long
#define uLL unsigned long long
#define debug(x) cout << "[" << x << "]" << endl
using namespace std;

const int mx = 1e5+10;
const int base = 131;
char s[mx], t[mx];
uLL hs[mx], ht[mx], a[mx], p[mx];
int n, m;

bool check(int x){
    int cnt = 0;
    for (int i = x; i <= n; i++)
        a[cnt++] = hs[i]-p[x]*hs[i-x];
    sort(a, a+cnt);
    for (int i = x; i <= m; i++){
        uLL k = ht[i]-p[x]*ht[i-x];
        if (binary_search(a, a+cnt, k)) return 0;
    }
    return 1;
}

int main(){
    p[0] = 1;
    scanf("%s%s", s+1, t+1);
    n = strlen(s+1), m = strlen(t+1);
    for (int i = 1; i <= n; i++) hs[i] = hs[i-1]*base+s[i]-'a';
    for (int i = 1; i <= m; i++) ht[i] = ht[i-1]*base+t[i]-'a';
    for (int i = 1; i <= max(n, m); i++) p[i] = p[i-1]*base;
    int l = 0, r = min(n, m);
    while (l <= r){
        int mid = (l+r)>>1;
        if (check(mid)) r = mid-1;
        else l = mid+1;
    }
    printf("%d\n", r);
    return 0;
}