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最新推荐文章于 2020-07-22 21:10:50 发布

转载最新推荐文章于 2020-07-22 21:10:50 发布 · 64 阅读

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原文链接：http://www.cnblogs.com/cjoieryl/p/8822988.html

本文介绍了一道关于字符串Hash和排序算法的题目解答过程，通过使用双Hash技巧及前后缀Hash值来快速比较子串，实现了高效的子串匹配算法。

题面

传送门

Sol

\(hash\)+排序

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;

template <class Int>
IL void Input(RG Int &x){
    RG int z = 1; RG char c = getchar(); x = 0;
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    x *= z;
}

const int maxn(3e4 + 5);
const int maxm(205);
const int seed1(1402);
const int seed2(1707);

int n, m, s, id[maxn], ans;
ll p1[2][maxn][maxm], p2[2][maxn][maxm], key1[maxn], key2[maxn];
char name[maxn][maxm];

IL int Cmp(RG int x, RG int y){
    return key1[x] == key1[y] ? key2[x] < key2[y] : key1[x] < key1[y];
}

int main(RG int argc, RG char* argv[]){
    Input(n), Input(m), Input(s);
    for(RG int i = 1; i <= n; ++i){
        scanf(" %s", name[i] + 1), id[i] = i;
        for(RG int j = 1; j <= m; ++j){
            p1[0][i][j] = p1[0][i][j - 1] * seed1 + name[i][j];
            p1[1][i][j] = p1[1][i][j - 1] * seed2 + name[i][j];
        }
        for(RG int j = m; j; --j){
            p2[0][i][j] = p2[0][i][j + 1] * seed1 + name[i][j];
            p2[1][i][j] = p2[1][i][j + 1] * seed2 + name[i][j];
        }
    }
    for(RG int j = 1; j <= m; ++j){
        for(RG int i = 1; i <= n; ++i){
            key1[i] = p1[0][i][j - 1] * seed1 + p2[0][i][j + 1] * seed2;
            key2[i] = p1[1][i][j - 1] * seed2 + p2[1][i][j + 1] * seed1;
        }
        sort(id + 1, id + n + 1, Cmp);
        for(RG int p = 2, num = 1, pre = id[1]; p <= n; ++p){
            RG int i = id[p];
            if(key1[i] == key1[pre] && key2[i] == key2[pre]) ans += num, ++num;
            else num = 1;
            pre = i;
        }
    }
    printf("%d\n", ans);
    return 0;
}

转载于:https://www.cnblogs.com/cjoieryl/p/8822988.html