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palindrome mouse

admin 11月 14, 2020 0

题目链接

题意

现在有一个字符串$s$,你从中可以找到多少个回文子串对，使其中一个是另一个的子串。

思路

在回文树上建出这个串，那么每个点所表示的回文串，包含它的祖先作为子串，另一部分还有所有fail指针指向的点，都是它的后缀。
但是要注意，有的时候这两个方向可能有重合的点，如$aaaa$。
往上找答案的时候加一个vis标记去记录是否已经加过这个点。

Code



using namespace std;
typedef long long ll;

const int maxn = 1e5 + 5;

int _, kase = 1;
char s[maxn];
ll ans = 0;

struct  {
    int next[maxn][26];
    int fail[maxn];
    int len[maxn];
    int S[maxn];
    int dp[maxn];
    bool vis[maxn];
    int last, n, p;


    int newNode(int l) {
        memset(next[p], 0, sizeof(next[p]));
        len[p] = l;
        dp[p] = 0;
        return p++;
    }

    void init() {
        ans = 0;
        n = last = p = 0;
        newNode(0);
        newNode(-1);
        S[n] = -1;
        fail[0] = 1;
    }

    int getFail(int x) {
        while (S[n - len[x] - 1] != S[n]) {
            x = fail[x];
        }
        return x;
    }

    void add(int c) {
        S[++n] = c;
        int cur = getFail(last);
        if (!next[cur][c]) {
            int now = newNode(len[cur] + 2);
            fail[now] = next[getFail(fail[cur])][c];
            next[cur][c] = now;
        }
        last = next[cur][c];
    }

    int jump(int x) {
        int cnt = 0;
        vis[x] = 1;
        while (fail[x] != 0 && fail[x] != 1 && !vis[fail[x]]) {
            x = fail[x];
            vis[x] = 1, ++cnt;
        }
        return cnt;
    }

    void clearJump(int x, int cnt) {
        vis[x] = 0;
        while (cnt--) {
            x = fail[x];
            vis[x] = 0;
        }
    }

    void dfs(int x, int fa) {
        int jp = jump(x);
        dp[x] = jp;
        if (x != 1 && x != 0 && fa != 0 && fa != 1) {
            dp[x] = dp[fa] + jp + 1;
        }
        ans += dp[x];
        for (int i = 0; i < 26; ++i) {
            if (next[x][i]) {
                dfs(next[x][i], x);
            }
        }
        clearJump(x, jp);
    }

    void build() {
        init();
        for (int i = 1; s[i]; i++) {
            add(s[i] - 'a');
        }
    }

} pam;


int main() {
    scanf("%d", &_);
    while (_--) {
        scanf("%s", s + 1);
        pam.build();
        printf("Case #%d: ", kase++);
        pam.dfs(1, 1);
//        printf("%lldn", ans);
        pam.dfs(0, 0);
        printf("%lldn", ans);
    }
    return 0;
}

甚至还可以在后缀树上建主席树。（待补

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 1e5 + 10, N = 1e5 + 10;
queue<int> q;
int rt[maxn], ls[maxn * 50], rs[maxn * 50], sum[maxn * 50], cnt;
#define mid (l + r) / 2
void up(int &o, int pre, int l, int r, int k) {
    o = ++cnt;
    sum[o] = sum[pre] + 1;
    ls[o] = ls[pre];
    rs[o] = rs[pre];
    if (l == r)
        return;
    if (k <= mid)
        up(ls[o], ls[pre], l, mid, k);
    else
        up(rs[o], rs[pre], mid+ 1, r, k);
}
int qu(int o, int l, int r, int k) {
    if (l == r)
        return sum[o];
    if (k <= mid)
        return qu(ls[o], l, mid, k);
    return qu(rs[o], mid + 1, r, k);
}
struct ptree{
    char s[maxn];
    int next[maxn][26],fail[maxn],cnt[maxn],len[maxn], d[maxn];
    int last,n,p;
    long long res;
    ll ans = 0;
    inline int newnode(int l){
        cnt[p]=0;
        len[p]=l;
        memset(next[p], 0, sizeof(next[p]));
        return p++;
    }
    inline void init(){
        n = last = p = ans = 0;
        newnode(0),newnode(-1);
        s[n]=-1;
        fail[0]=1;
    }
    inline int FL(int x){
        while(s[n-len[x]-1]!=s[n]) x=fail[x];
        return x;
    }
    void add(char c){
        c-='a';
        s[++n]=c;
        int cur=FL(last);
        if(!next[cur][c]){
            int now=newnode(len[cur]+2);
            cnt[now] = 1;
            rt[now] = rt[cur];
            int FF = next[FL(fail[cur])][c];
            fail[now]=next[FL(fail[cur])][c];
            next[cur][c]=now;
            while (FF > 1) {
                if (qu(rt[now], 1, N, FF))
                    break;
                up(rt[now], rt[now], 1, N, FF);
                FF = fail[FF];
            }
            up(rt[now], rt[now], 1, N, now);
            ans += sum[rt[now]] - 1;
        }
        last=next[cur][c];
    }
} p;
char s[maxn];
int main(){
    int T, Case = 0;
    scanf("%d", &T);
    while (T--) {
        scanf("%s",s);
        p.init();
        cnt = 0;
        for (int i = 0; s[i]; i++)
            p.add(s[i]);
        printf("Case #%d: %lldn", ++Case, p.ans);
    }
}