BZOJ4566: [Haoi2016]找相同字符(后缀自动机)

weixin_30399155

于 2018-12-14 10:57:00 发布

阅读量61

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zwfymqz/p/10118217.html

本文深入探讨了字符串匹配算法SAM（Suffix Array Machine）的核心概念，通过实战案例展示如何使用SAM解决复杂字符串匹配问题，包括前缀和后缀的枚举统计、最大值维护等关键步骤。

题意

题目链接

Sol

直接在SAM上乱搞

枚举前缀，用SAM统计可以匹配的后缀，具体在匹配的时候维护和当前节点能匹配的最大值

然后再把parent树上的点的贡献也统计上，~~这部分可以爆跳parent树~~(假的，因为这题数据随机)，也可以直接树形dp一波记下每个点被统计的次数

#include<bits/stdc++.h>
#define LL long long 
using namespace std;
const int MAXN = 4e5 + 10;
int N1, N2;
char a[MAXN], b[MAXN];
// struct SAM {
    int fa[MAXN], ch[MAXN][26], len[MAXN], siz[MAXN], tot, las, root, f[MAXN], g[MAXN];
    vector<int> v[MAXN];
    // SAM() {
        // root = las = tot = 1;
    // }
    void insert(int x) {
        int now = ++tot, pre = las; las = now; siz[now] = 1;
        len[now] = len[pre] + 1;
        for(; pre && !ch[pre][x]; pre = fa[pre]) ch[pre][x] = now;
        if(!pre) fa[now] = root;
        else {
            int q = ch[pre][x];
            if(len[pre] + 1 == len[q]) fa[now] = q;
            else {
                int ns = ++tot; fa[ns] = fa[q]; len[ns] = len[pre] + 1;
                memcpy(ch[ns], ch[q], sizeof(ch[q]));
                fa[q] = fa[now] = ns;
                for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = ns;
            }
        }
    }
    void Build() {
        for(int i = 2; i <= tot; i++) v[fa[i]].push_back(i);
    }
    void dfs(int x, int opt) {
        for(int i = 0; i < v[x].size(); i++) {
            int to = v[x][i];
            dfs(to, opt); 
            if(opt == 1) siz[x] += siz[to]; 
            if(opt == 2) f[x] += f[to] + g[to];
        }
    }
// }sam;
int main() {
    root = las = tot = 1;
    scanf("%s%s", a + 1, b + 1);
    N1 = strlen(a + 1); N2 = strlen(b + 1);
    for(int i = 1; i <= N1; i++) insert(a[i] - 'a');
    Build(); dfs(root, 1);
    int now = root, cur = 0; LL ans = 0;
    for(int i = 1; i <= N2; i++) {
        int x = b[i] - 'a';
        if(ch[now][x]) now = ch[now][x], cur++;
        else {
            while(!ch[now][x] && now) now = fa[now];
            if(!now) now = 1, cur = 0;
            else cur = len[now] + 1, now = ch[now][x];
        }
        ans += 1ll * (cur - len[fa[now]]) * siz[now];
        g[now]++;
        //int tmp = fa[now];
        //while(tmp != 1) ans += (len[tmp] - len[fa[tmp]]) * siz[tmp], tmp = fa[tmp];
    }
    dfs(root, 2);
    for(int i = 1; i <= tot; i++) ans += 1ll * (len[i] - len[fa[i]]) * siz[i] * f[i];
    cout << ans;
    return 0;
}
/*
aa
aa

acb
abc

ababababba
abbababab
*/

转载于:https://www.cnblogs.com/zwfymqz/p/10118217.html