2018 ACM-ICPC 中国大学生程序设计竞赛线上赛 H-Rock Paper Scissors Lizard Spock.

最新推荐文章于 2021-01-27 10:18:30 发布

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本文介绍了一种使用快速傅里叶变换（FFT）解决字符串匹配问题的方法，通过将字符串转换为数值序列并利用FFT进行高效卷积计算，实现了字符串的匹配与计数。

传送门：Rock Paper Scissors Lizard Spock.

这题还有一个弱化版：GYM 101667-H

分析：石头剪刀布加强版。

把短的字符串跟长的字符串从某一位置进行匹配，问最多可以匹配多少位。

把其中一个串反转，对每一种字符单独计算。

字符匹配的过程看做卷积和，则当前要匹配的字符对应1，否则为0。

用FFT快速求解，最后取最大值。

代码如下：

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;

const int maxn = 4e6+10;
const double PI = acos(-1.0);

struct complex{
     double r,i;
     complex (double r=0.0, double i=0.0):r(r),i(i){}
     complex operator +(const complex &rhs) {return complex(r+rhs.r,i+rhs.i); }
     complex operator -(const complex &rhs) {return complex(r-rhs.r,i-rhs.i); }
     complex operator *(const complex &rhs) {return complex(r*rhs.r-i*rhs.i,r*rhs.i+i*rhs.r); }
};

void change(complex y[], int len) {
    int i,j,k;
    for (int i=1, j = len/2; i<len-1; i++) {
        if (i<j) swap(y[i],y[j]);
        k = len/2;
        while (j>=k) {
            j-=k;
            k/=2;
        }
        if (j<k) j+=k;
    }
}

void fft(complex y[], int len, int on) {
    change(y,len);
    for (int h=2; h<=len; h<<=1) {
        complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));
        for (int j=0; j<len; j+=h) {
            complex w(1,0);
            for (int k=j; k<j+h/2; k++) {
                complex u = y[k];
                complex t = w*y[k+h/2];
                y[k] = u+t;
                y[k+h/2] = u-t;
                w = w*wn;
            }
        }
    }

    if (on == -1) {
        for (int i=0; i<len; i++) y[i].r /= len;
    }
}

complex x1[maxn],x2[maxn];
char a[maxn/2],b[maxn/2],a1[maxn/2];
int sum[maxn],ans[maxn];
int n1,n2;

void calc(char c){
   int len = 1;
   while (len<n1*2 || len<n2*2) len<<=1;
   for (int i=0; i<n1; i++) x1[i] = complex(a[i]==c,0);
   for (int i=n1; i<len; i++) x1[i] = complex(0,0);
   for (int i=0; i<n2; i++) x2[i] = complex(b[i]==c,0);
   for (int i=n2; i<len; i++) x2[i] = complex(0,0);
   fft(x1,len,1);
   fft(x2,len,1);
   for (int i=0; i<len; i++) x1[i] = x1[i]*x2[i];
   fft(x1,len,-1);
   for (int i=0; i<len; i++) sum[i] = (int)(x1[i].r+0.5);
   len = n1+n2-1;
   for (int i=0; i<len; i++) ans[i] += sum[i];
}

int main(){
    while (scanf("%s%s",a1,b)==2) {
        n1 = strlen(a1);
        n2 = strlen(b);
        for (int i=0; i<=n1+n2-1; i++) ans[i] = 0;
        reverse(b,b+n2);
        for (int i=0; i<n1; i++) {
            if (a1[i] == 'P') a[i] = 'S';
            else if (a1[i] == 'R') a[i] = 'P';
            else if (a1[i] == 'L') a[i] = 'R';
            else if (a1[i] == 'S') a[i] = 'K';
            else if (a1[i] == 'K') a[i] = 'L';
        }
        calc('S');  calc('R'); calc('P');  calc('L'); calc('K');

        for (int i=0; i<n1; i++) {
            if (a1[i] == 'P') a[i] = 'L';
            else if (a1[i] == 'R') a[i] = 'K';
            else if (a1[i] == 'L') a[i] = 'S';
            else if (a1[i] == 'S') a[i] = 'R';
            else if (a1[i] == 'K') a[i] = 'P';
        }
        calc('S');  calc('R'); calc('P');  calc('L'); calc('K');

        int total = 0;
        for (int i=n2-1; i<n1+n2-1; i++) total = max(ans[i],total);
        printf("%d\n",total);
    }
    return 0;
}