题面

HiHocoder1415

Poj2774

Sol

都是求最长公共子串，\(hihocoder\)上讲的很清楚

把两个串拼在一起，中间用一个特殊字符隔开

那么答案就是排序后相邻两个不同串的后缀的\(height\)

为什么呢？

如果答案为不相邻的两个后缀的前缀，计算它们最长前缀时必定要跨越过这些中间\(height\)值，也就是选相邻的两个一定要比不选相邻的两个更优

# 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;

const int _(200010);

IL ll Read(){

    RG char c = getchar(); RG ll x = 0, z = 1;

    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);

    return x * z;

}

int n, a[_], sa[_], rk[_], y[_], h[_], height[_], t[_], ans;

char A[_], B[_];

IL bool Cmp(RG int i, RG int j, RG int k){  return y[i] == y[j] && y[i + k] == y[j + k];  }

IL void Sort(){

    RG int m = 27;

    for(RG int i = 1; i <= n; ++i) ++t[rk[i] = a[i]];

    for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];

    for(RG int i = n; i; --i) sa[t[rk[i]]--] = i;

    for(RG int k = 1; k <= n; k <<= 1){

        RG int l = 0;

        for(RG int i = n - k + 1; i <= n; ++i) y[++l] = i;

        for(RG int i = 1; i <= n; ++i) if(sa[i] > k) y[++l] = sa[i] - k;

        for(RG int i = 0; i <= m; ++i) t[i] = 0;

        for(RG int i = 1; i <= n; ++i) ++t[rk[y[i]]];

        for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];

        for(RG int i = n; i; --i) sa[t[rk[y[i]]]--] = y[i];

        swap(rk, y); rk[sa[1]] = l = 1;

        for(RG int i = 2; i <= n; ++i) rk[sa[i]] = Cmp(sa[i - 1], sa[i], k) ? l : ++l;

        if(l >= n) break; m = l;

    }

    for(RG int i = 1; i <= n; ++i){

        h[i] = max(0, h[i - 1] - 1);

        if(rk[i] == 1) continue;

        while(a[i + h[i]] == a[sa[rk[i] - 1] + h[i]]) ++h[i];

    }

    for(RG int i = 1; i <= n; ++i) height[i] = h[sa[i]];

}

IL bool Diff(RG int a, RG int b, RG int c){  return (a <= c && b > c) || (a > c && b <= c);  }

int main(RG int argc, RG char* argv[]){

	scanf(" %s %s", A, B); RG int l1 = strlen(A), l2 = strlen(B);

	for(RG int i = 0; i < l1; ++i) a[++n] = A[i] - 'a' + 1; a[++n] = 27;

	for(RG int i = 0; i < l2; ++i) a[++n] = B[i] - 'a' + 1;

    Sort();

	for(RG int i = 2; i <= n; ++i)

		if(Diff(sa[i - 1], sa[i], l1)) ans = max(ans, height[i]);

	printf("%d\n", ans);

    return 0;

}