Gym10198-Mediocre String Problem-2018南京ICPC现场赛

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Problem:传送门

Portal

 原题目描述在最下面。


Solution:

  • 二分+\(hash\)+\(manacher\)
  • 题意要我们在s串中找一个子串,和在t串中找一个前缀串,组合成回文串,但是串1长度要大于串2的长度。
  • 其实就是要找三个串,在s中找连续的串a和串b,在t中找一个前缀串c
  • 满足b是回文串且长度至少为1,a串和c串对称
  • 到这一步思路就十分明确了,可以用\(manacher\)预处理出以\(i\)点为左端点的串是回文串的个数,差分一下就可以了。
  • 然后就很简单了,先把t串反转。
  • 枚举\(i\)\(i\)表示a串的末尾,则\(i+1\)就是b串的起点,然后求s串[0,i]的子串和反转后t串的最长公共后缀,这一步就直接二分+\(hash\)就可以了。累计贡献\(len_{最长公共后缀}\times pre[i+1]\)

AC_Code:

#include<bits/stdc++.h>
#define lson rt<<1
#define rson rt<<1|1
#define fi first
#define se second
#define lowbit(x) (x&(-(x)))
#define mme(a,b) memset((a),(b),sizeof((a))) 
#define fuck(x) cout<<"* "<<x<<"\n"
#define iis std::ios::sync_with_stdio(false)
using namespace std;
typedef long long LL;
const int N = 1e6+7;
const int MXN = 1e6+7;
const long long MOD = 2078526727;
const long long BASE = 1572872831;
LL pw[MXN], hs1[MXN], hs2[MXN];
int n, m;
char ar[N<<1], br[N<<1], t[N];
int p[N<<1], pre[N];
void manacher(){
  int Len = strlen(ar), id = 0, ans_id = 0, right = 0;
  memset(p, 0, sizeof(p));
  memset(br, 0, sizeof(br));
  for(int i = Len; i >= 0; --i){
    ar[2 * i + 1] = '#';
    ar[2 * i + 2] = ar[i];
  }
  ar[0] = '*';p[0] = p[1] = 1;
  for(int i = 2; i < 2 * Len + 1; ++i){
    if(i < right)p[i] = min(p[2 * id - i], right - i);
    else p[i] = 1;
    while(i - p[i] >= 0 && ar[i + p[i]] == ar[i - p[i]])p[i]++;
    if(p[ans_id] < p[i])ans_id = i;
    if(i + p[i] > right){
      id =i;
      right = i + p[i];
    }
  }
}
LL get(LL *a, int l, int r) {
    return ((a[r]-a[l-1]*pw[r-l+1])%MOD+MOD)%MOD;
}
bool ok(int len, int P, int lent) {
    if(get(hs1, P-len+1, P) == get(hs2, lent-len+1, lent)) return true;
    return false;
}
int main() {
    pw[0] = 1;
    for(int i = 1; i < 1000003; ++i) pw[i] = pw[i-1] * BASE % MOD;
    scanf("%s%s", ar, t);
    int len = strlen(ar);
    hs1[0] = hs2[0] = 1;
    for(int i = 0; i < len; ++i) {
        hs1[i+1] = (hs1[i]*BASE + ar[i]) % MOD;
    }
    manacher();
    for(int i = 2; i < 2 * len + 1; ++ i) {
        pre[(i-p[i]+2)/2-1] ++;
        pre[i/2] --;
        //printf("%d %d %d %d\n", i, p[i], (i-p[i]+2)/2-1, i/2);
    }
    for(int i = 1; i < len; ++i) {
        pre[i] += pre[i-1];
    }
    /*for(int i = 0; i < len; ++i) {
        printf("%d ", pre[i]);
    }
    printf("\n");*/
    int lent = strlen(t);
    reverse(t, t + lent);
    for(int i = 0; i < lent; ++i) {
        hs2[i+1] = (hs2[i]*BASE + t[i]) % MOD;
    }
    int k = 0;
    for(int i = 2; i < 2*len-1; ++i) {
        if(i%2 == 0) ar[k++] = ar[i];
    }
    LL res = 0;
    for(int i = 0; i < len - 1; ++i) {
        //二分LCP,[?,i],[1,?]
        if(ar[i] != t[lent-1]) continue;
        int l = 1, r = min(lent, i+1), mid, ans = 0;
        while(r >= l) {
            mid = (l+r)>>1;
            if(ok(mid, i+1, lent)) ans = mid, l = mid+1;
            else r = mid-1;
        }
        res += (LL)ans * pre[i+1];
        //printf("%d %d\n", i, ans);
    }
    printf("%lld\n", res);
    return 0;
}

Problem Description:

在这里插入图片描述

posted @ 2018-11-21 13:40  Cwolf9  阅读(168)  评论(1编辑  收藏

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