CodeForces 149E Martian Strings exkmp

Martian Strings

题解：

对于询问串， 我们可以从前往后先跑一遍exkmp。

然后在倒过来，从后往前跑一遍exkmp。

我们就可以记录下 对于每个正向匹配来说，最左边的点在哪里。

对于每个反向匹配来说，最右边的点在哪里。

然后判断可不可以构成这个串就好了。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 2e5 + 100;
int n, z[N];
char s[N];
void init(int n){
    z[0] = n;
    int j = 1, k;
    for(int i = 1; i < n; i = k){
        if(j < i) j = i;
        while(j < n && s[j] == s[j-i]) j++;
        z[i] = j-i;
        k = i+1;
        while(k + z[k-i] < j)
            z[k] = z[k-i],k++;
    }
}
char ss[N], t[N];
int l[N], r[N];
int q;
int main(){
    scanf("%s", ss);
    int lens = strlen(ss);
    scanf("%d", &q);
    int ans = 0;
    while(q--){
        scanf("%s", t);
        int len2 = strlen(t);
        for(int i = 1; i <= len2; ++i){
            l[i] = lens + 1;
            r[i] = 0;
        }
        strcpy(s, t);
        s[len2] = '@';
        strcpy(s+1+len2, ss);
        init(lens + 1 + len2);
        for(int i = 1; i <= lens; ++i){
            int t = z[len2+i];
            l[t] = min(l[t], i);
        }
        for(int i = len2-1; i >= 1; --i)
            l[i] = min(l[i], l[i+1]);
        reverse(s, s+len2);
        reverse(s+len2+1, s+lens+1+len2);
        init(lens + 1 + len2);
        for(int i = 1; i <= lens; ++i){
            int t = z[len2+i];
            r[t] = max(r[t], lens-i+1);
        }
        for(int i = len2-1; i >= 1; --i)
            r[i] = max(r[i], r[i+1]);
        for(int i = 1; i < len2; ++i){
            if(l[i] + len2-1 <= r[len2-i]) {
                ans++;
                break;
            }
        }
    }
    cout << ans << endl;
    return 0;
}