ACM-ICPC2018南京赛区 Mediocre String Problem

Mediocre String Problem

题解：

很容易想到将第一个串反过来，然后对于s串的每个位置可以求出t的前缀和它匹配了多少个（EXKMP 或者 二分+hash）。

然后剩下的就是要处理以某个位置为结束的回文串有多少个（manacher + 差分），因为要求s串选取的要多一点。
这道题是个痛啊。。。当时的金牌题，不会EXKMP可以用二分+字符串hash啊，比赛前的暑假还写过，比赛时就没想到，还以为KMP可以搞出这个东西，

然后就三个人一起自闭地调KMP，说到底还是菜呀。

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "\n";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head

const int N = 1e6 + 10;
int p[N*2], cnt[N];
char s[N], t[N];
int nxt[N], ex[N];
void GETNEXT(char *str) {
    int i = 0, j, po, len=strlen(str);
    nxt[0] = len;
    while(str[i] == str[i+1] && i+1 < len) i++;
    nxt[1] = i;
    po = 1;
    for(i = 2; i < len; i++) {
        if(nxt[i-po] + i < nxt[po] + po)
        nxt[i] = nxt[i-po];
        else {
            j=nxt[po] + po - i;
            if(j < 0) j = 0;
            while(i + j < len && str[j] == str[j+i])
            j++;
            nxt[i] = j;
            po = i;
        }
    }
}
void EXKMP(char *s1,char *s2)
{
    int i = 0, j, po, len = strlen(s1), l2=strlen(s2);
    GETNEXT(s2);
    while(s1[i] == s2[i] && i < l2 && i < len) i++;
    ex[0] = i;
    po = 0;
    for(i = 1; i < len; i++)
    {
        if(nxt[i-po] + i < ex[po] + po) ex[i]=nxt[i-po];
        else {
            j = ex[po] + po - i;
            if(j < 0) j = 0;
            while(i + j < len && j < l2 && s1[j+i] == s2[j]) j++;
            ex[i] = j;
            po = i;
        }
    }
}
void manacher(char *s) {
    string t = "$#";
    int n = strlen(s);
    for (int i = 0; i < n; ++i) {
        t += s[i];
        t += '#';
    }
    int mx = 0, id = 0, resl = 0, resc = 0;
    for (int i = 1; i < t.size(); ++i) {
        p[i] = mx > i ? min(p[2*id-i], mx-i) : 1;
        while(t[i+p[i]] == t[i-p[i]]) ++p[i];
        if(mx < i+p[i]) mx = i+p[i], id = i;
        if(resl < p[i]) resl = p[i], resc = i;
    }
    for (int i = 1; i < t.size(); ++i) {
        if(p[i] == 1 && t[i] == '#') continue;
        int l, r;
        if(p[i]&1) {
            l = (i-1)/2;
            int d = (p[i]-1)/2;
            r = l+d;
        }
        else {
            l = (i-2)/2;
            int d = p[i]/2;
            r = l+d;
        }
        cnt[l]++, cnt[r]--;
    }
    for (int i = 1; i < n; ++i) cnt[i] += cnt[i-1];
}

int main() {
    scanf("%s", s);
    scanf("%s", t);
    int n = strlen(s);
    for (int i = 0, j = n-1; i < j; ++i, --j) {
        swap(s[i], s[j]);
    }
    manacher(s);
    EXKMP(s, t);
    LL ans = 0;
    for (int i = 1; i < n; ++i) {
        ans += 1LL * ex[i] * cnt[i-1];
    }
    printf("%lld\n", ans);
    return 0;
}