P3900 [湖南集训]图样图森破

P3900 [湖南集训]图样图森破

链接

分析：

　　感觉像个暴力。

　　可以枚举回文串的回文中心，即枚举一个串，枚举一个串的位置作为回文中心，然后求出这个串内的回文串的长度。

　　此时如果回文串两端都没有到这个串的端点，那么以这个点作为回文中心的长度就直接算出来了。

　　如果回文串的长度刚好是这个串的长度，那么INF。

　　如果回文串一侧到了端点，那么枚举所有串，看看能否加到另一侧，来构成回文串。此过程记忆化搜索即可。

　　复杂度$O(nL \log nL)$

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;

inline int read() {
    int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
    for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
}

const int N = ;
int s[N], st[N], en[N], rnk[N], ht[N], sa[N], f[][N], Log[N], t1[N], t2[N], c[N], bel[N], dp[N][];
bool vis[N][];
char tmp[N];
int n, m;

void getsa() {
    int *x = t1, *y = t2, m = , i;
    for (i = ; i <= m; ++i) c[i] = ;
    for (i = ; i <= n; ++i) x[i] = s[i], c[x[i]] ++;
    for (i = ; i <= m; ++i) c[i] += c[i - ];
    for (i = n; i >= ; --i) sa[c[x[i]] --] = i;
    for (int k = ; k <= n; k <<= ) {
        int p = ;
        for (i = n - k + ; i <= n; ++i) y[++p] = i;
        for (i = ; i <= n; ++i) if (sa[i] > k) y[++p] = sa[i] - k;
        for (i = ; i <= m; ++i) c[i] = ;
        for (i = ; i <= n; ++i) c[x[y[i]]] ++;
        for (i = ; i <= m; ++i) c[i] += c[i - ];
        for (i = n; i >= ; --i) sa[c[x[y[i]]] --] = y[i];
        swap(x, y);
        p = ;
        x[sa[]] = ;
        for (i = ; i <= n; ++i)
            x[sa[i]] = (y[sa[i]] == y[sa[i - ]] && y[sa[i] + k] == y[sa[i - ] + k]) ? p -  : p ++;
        if (p > n) break;
        m = p;
    }
    for (int i = ; i <= n; ++i) rnk[sa[i]] = i;
    int k = ;
    ht[] = ;
    for (int i = ; i <= n; ++i) {
        if (rnk[i] == ) continue;
        if (k) k --;
        int j = sa[rnk[i] - ];
        while (i + k <= n && j + k <= n && s[i + k] == s[j + k]) k ++;
        ht[rnk[i]] = k;
    }
    Log[] = -;
    for (int i = ; i <= n; ++i) Log[i] = Log[i >> ] + ;
    for (int i = ; i <= n; ++i) f[][i] = ht[i];
    for (int j = ; j <= Log[n]; ++j)
        for (int i = ; i + ( << j) -  <= n; ++i)
            f[j][i] = min(f[j - ][i], f[j - ][i + ( << (j - ))]);
}
int query(int l,int r) {
    if (l == r) return 1e9;
    l = rnk[l], r = rnk[r];
    if (l > r) swap(l, r);
    l ++;
    int k = Log[r - l + ];
    return min(f[k][l], f[k][r - ( << k) + ]);
}
int getlcp(int x,int y) {
    return min(query(x, n - y + ), min(en[bel[x]] - x + , y - st[bel[y]] + ));
}
void End() {
    puts("Infinity"); exit();
}
int dfs(int x,int t) {
    if (vis[x][t]) End();
    if (dp[x][t]) return dp[x][t];
    vis[x][t] = ;
    if (!t) {
        for (int i = ; i <= m; ++i) {
            int k = getlcp(x, en[i]), l = en[i] - k + , r = x + k - ;
            if (r != en[bel[x]] && l != st[i]) dp[x][t] = max(dp[x][t], k * );
            else if (r == en[bel[x]] && l == st[i]) End();
            else if (r == en[bel[x]]) dp[x][t] = max(dp[x][t], k *  + dfs(l - , ));
            else dp[x][t] = max(dp[x][t], k *  + dfs(r + , ));
        }
    }
    else {
        for (int i = ; i <= m; ++i) {
            int k = getlcp(st[i], x), l = x - k + , r = st[i] + k - ;
            if (l != st[bel[x]] && r != en[i]) dp[x][t] = max(dp[x][t], k * );
            else if (l == st[bel[x]] && r == en[i]) End();
            else if (l == st[bel[x]]) dp[x][t] = max(dp[x][t], k *  + dfs(r + , ));
            else dp[x][t] = max(dp[x][t], k *  + dfs(l - , ));
        }
    }
    vis[x][t] = ;
    return dp[x][t];
}
int main() {
    m = read();
    for (int i = ; i <= m; ++i) {
        scanf("%s", tmp + );
        int len = strlen(tmp + );
        st[i] = n + ;
        for (int j = ; j <= len; ++j) s[++n] = tmp[j] - 'a' + , bel[n] = i;
        en[i] = n;
    }
    for (int i = ; i <= n; ++i) s[i + n] = s[n - i + ];
    n <<= ;
    getsa();
    int ans = ;
    for (int i = ; i <= m; ++i)
        ans = max(ans, max(dfs(st[i], ), dfs(en[i], )));
    for (int i = ; i <= m; ++i) {
        for (int j = st[i]; j <= en[i]; ++j) {
            int k = getlcp(j, j), l = j - k + , r = j + k - ;
            if (l != st[i] && r != en[i]) ans = max(ans, k *  - );
            else if (l == st[i] && r == en[i]) End();
            else if (l == st[i]) ans = max(ans, k *  -  + dfs(r + , ));
            else ans = max(ans, k *  -  + dfs(l - , ));
        }
        for (int j = st[i]; j < en[i]; ++j) {
            int k = getlcp(j + , j), l = j - k + , r = j + k; // r = j + 1 + k - 1 !!!
            if (l != st[i] && r != en[i]) ans = max(ans, k * );
            else if (l == st[i] && r == en[i]) End();
            else if (l == st[i]) ans = max(ans, k *  + dfs(r + , ));
            else ans = max(ans, k *  + dfs(l - , ));
        }
    }
    cout << ans;
    return ;
}