Rock Paper Scissors Lizard Spock

题意：

Rock Paper Scissors Lizard Spock

有五种手势，类似于石头剪刀布，有两个串\(s, t\),由这五种手势组成，从某个位置开始匹配，如果\(t_i\)能赢\(s_j\)得一分，求一个\(pos(0\le pos \le len(s)-len(t))\)，使得得分最多

Solution：

将上图记为\(G\)，如果\(op_1\)可以赢\(op_2\)，则\(G(op_1, op_2)=1\),

枚举可以得分的手势,假设当前手势为\(op\),
\[ \begin{aligned} t'_i &=(reverse\_t_{i} == op ? 1 : 0) \\ s'_i &= G[op][s_i] \end{aligned} \]

将\(t',s'\)做卷积，累加每次的结果，再遍历一遍匹配的起始位置，取最大值

#include <cstdio>
#include <stack>
#include <set>
#include <cmath>
#include <map>
#include <time.h>
#include <vector>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <memory.h>
#include <cstdlib>
#include <queue>
#include <iomanip>
#include <unordered_map>
#define P pair<int, int>
#define LL long long
#define LD long double
#define PLL pair<LL, LL>
#define mset(a, b) memset(a, b, sizeof(a))
#define rep(i, a, b) for (int i = a; i < b; i++)
#define PI acos(-1.0)
#define random(x) rand() % x
#define debug(x) cout << #x << " " << x << "\n"
using namespace std;
const int inf = 0x3f3f3f3f;
const LL __64inf = 0x3f3f3f3f3f3f3f3f;
#ifdef DEBUG
const int MAX = 2e3 + 50;
#else
const int MAX = 1e6 + 50;
#endif
const int mod = 1e9 + 7;

#include <complex>
namespace FFT
{
    int rev[MAX * 4];
    int cnt[MAX * 4];
    void fft(complex<double> *a, int n, int inv){
        int bit = 0;
        while((1<<bit) < n) bit++;
        for(int i = 0; i < n; i++){
            rev[i] = (rev[i>>1]>>1) | ((i & 1) << (bit-1));
            if(i < rev[i])
                swap(a[i], a[rev[i]]);
        }
        for(int mid = 1; mid < n; mid <<= 1){
            complex<double> wn(cos(PI/mid), inv*sin(PI/mid));
            for(int i = 0; i < n; i += (mid<<1)){
                complex<double> w(1, 0);
                for(int j = 0; j < mid; j++, w*=wn){
                    complex<double> u = a[i+j], t = w * a[i+j+mid];
                    a[i+j] = u + t, a[i+j+mid] = u-t;
                }
            }
        }
    }
};
int N;
char s[MAX], t[MAX], rev_t[MAX];
complex<double> a[MAX*4], b[MAX*4];
int mark[100][100];
void init(){
    mark['S']['P'] = 1;
    mark['P']['R'] = 1;
    mark['R']['L'] = 1;
    mark['L']['K'] = 1;
    mark['K']['S'] = 1;
    mark['S']['L'] = 1;
    mark['L']['P'] = 1;
    mark['P']['K'] = 1;
    mark['K']['R'] = 1;
    mark['R']['S'] = 1;
}
void solve(char op, int n, int  m){
    for(int i = 0; i < n; i++) 
        a[i] = complex<double>(mark[op][s[i]], 0);
    for(int i = n; i < N; i++) 
        a[i] = complex<double>(0, 0);
    for(int i = 0; i < m; i++)
        b[i] = complex<double>(rev_t[i]==op?1:0, 0);
    for(int i = m; i < N; i++)
        b[i] = complex<double>(0, 0);
;
    FFT::fft(a, N, 1); FFT::fft(b, N, 1);
    for(int i = 0; i < N;i++)
        a[i] *= b[i];
    FFT::fft(a, N, -1);
    for(int i = m-1; i <= n-1; i++)
        FFT::cnt[i] += (int)(a[i].real() / N + 0.5);
    return;
}
int main(){
#ifdef DEBUG
    freopen("in", "r", stdin);
#endif
    init();
    scanf("%s%s", s, t);
    int n = strlen(s), m = strlen(t);
    N = 1;
    while(N <= n+m) N <<= 1;
    for(int i = 0; i < m; i++)  rev_t[i] = t[m-i-1];
    vector<char> ops({'R','P', 'S', 'K', 'L'});
    for(auto op : ops){
        solve(op, n, m);
    }
    int ans = 0;
    for(int i = m-1; i <= n-1; i++){
        ans = max(ans, FFT::cnt[i]);
        // printf("%d ", FFT::cnt[i]);
    }
    // puts("");
    printf("%d\n", ans);
    return 0;
}

来源：https://www.cnblogs.com/sidolarry/p/12295313.html

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