ZR # 955 Origami

ZR # 955 Origami

solution:

Can then be found origami folded into the upper portion is not actually used, and the following because he must correspond to the same position, only affect the color of the answer and the bottom of each location coordinate range. Therefore, we need only consider the bottom can be, that we can put the equivalent of origami paper cutter, remove that part of the smaller each time.
A hash great palindromic child maintenance each column and each row of strings, recorded at the maximum rows and columns can be multiplied.

CODE:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>

using namespace std;

#define LL long long
const int N = 2e6 + 50;

char s[N],tt[N];
string p[N];
int HL[N],WL[N],m,n;
int HR[N],WR[N],t[N],f[N];

int manacher(int *p, char *ss, int Len) {
    int len = 0; s[0] = '(';
    for(int i = 1 ; i <= Len ; ++ i) {
        s[++ len] = '#';
        s[++ len] = ss[i];
    }
    s[++ len] = '#', s[++ len] = ')';
    int mx = 0, id = 0;
    for(int i = 1 ; i < len ; ++ i) {
        if(i < mx) p[i] = min(mx - i, p[2 * id - i]);
        else p[i] = 1;
        while(s[i + p[i]] == s[i - p[i]]) ++ p[i];
        if(i + p[i] > mx) mx = i + p[i], id = i;
    }
    return len;
}

int main() {
    scanf("%d%d",&n,&m);
    for(int i = 1 ; i < n  ; i++)
        HL[i] = HR[i] = 1; 
    for(int i = 1 ; i < m ; i++)
        WL[i] = WR[i] = 1;
    for(int i = 1 ; i <= n ; i++) 
        cin >> p[i];
    for(int i = 1 ; i <= n ; i++) {
        int len = m;
        for(int j = 1 ; j <= len ; j++) 
            tt[j] = p[i][j - 1];
        int Len = manacher(t, tt, m) - 3;
        int mx = 1;
        for(int j = 3 ; j <= Len ; j += 2) {
            if(j - t[j] + 1 <= mx) mx = j;
            else WL[j >> 1] = 0;
        }
        int mn = Len + 2;
        for(int j = Len ; j >= 3 ; j -= 2) {
            if(j + t[j] - 1 >= mn) mn = j;
            else WR[j >> 1] = 0;
        }
    }
    LL ansW = 0;
    int sum = 0;
    WL[0] = 1, WR[m] = 1;
    for(int i = 0 ; i <= m ; i++) sum += WR[i];
    for(int i = 0 ; i <= m ; i++) {
        if(WR[i]) -- sum;
        if(WL[i]) ansW += sum;
    }
    swap(n, m);
    for(int i = 1 ; i <= n ; i++) {
        int len = m;
        for(int j = 1 ; j <= len ; j++) 
            tt[j] = p[j][i - 1];
        int Len = manacher(t, tt, m) - 3;
        int mx = 1;
        for(int j = 3 ; j <= Len ; j += 2) {
            if(j - t[j] + 1 <= mx) mx = j;
            else HL[j >> 1] = 0;
        }
        int mn = Len + 2;
        for(int j = Len ; j >= 3 ; j -= 2) {
            if(j + t[j] - 1 >= mn) mn = j;
            else HR[j >> 1] = 0;
        }
    }
    LL ansH = 0;
    sum = 0;
    HL[0] = 1, HR[m] = 1;
    for(int i = 0 ; i <= m ; i++) sum += HR[i];
    for(int i = 0 ; i <= m ; i++) {
        if(HR[i]) -- sum;
        if(HL[i]) ansH += sum;
    }
    printf("%lld\n", ansW * ansH);
    //system("pause");
    return 0;   
}