ZR#955 折纸

ZR#955 折纸

解法:

可以发现折纸之后被折到上面的部分实际上是没有用的，因为他和下面对应位置一定是一样的，而影响答案的只有每个位置的颜色和最底层的坐标范围。因此，我们只需要考虑最底层即可，即我们可以把折纸等效为裁纸，每次去掉较小的那一部分。
用哈希维护每一列和每一行的极大回文子串，记录一下行与列的最大值相乘即可。

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;   
}

标签：int,mn,955,Len,HR,折纸,WR,include,ZR
来源： https://www.cnblogs.com/Repulser/p/11448968.html