2018 ACM-ICPC 中國大學生程序設計競賽線上賽：https://www.jisuanke.com/contest/1227
題目鏈接：https://nanti.jisuanke.com/t/26219
Rock Paper Scissors Lizard Spock
Description:
Didi is a curious baby. One day, she finds a curious game, which named Rock Paper Scissors Lizard Spock.
The game is an upgraded version of the game named Rock, Paper, Scissors. Each player chooses an option . And then those players show their choices that was previously hidden at the same time. If the winner defeats the others, she gets a point.
The rules are as follows.
Scissors cuts Paper
Paper covers Rock
Rock crushes Lizard
Lizard poisons Spock
Spock smashes Scissors
Scissors decapitates Lizard
Lizard eats Paper
Paper disproves Spock
Spock vaporizes Rock
(and as it always has) Rock crushes Scissors.
(this pic is from baike.baidu.com)
But Didi is a little silly, she always loses the game. In order to keep her calm, her friend Tangtang writes down the order on a list and show it to her. Didi also writes down her order on another list, like
.
(Rock-R Paper-P Scissors-S Lizard-L Spock-K)
However, Didi may skip some her friends' choices to find the position to get the most winning points of the game, like
Can you help Didi find the max points she can get?
Input:
The first line contains the list of choices of Didi's friend, the second line contains the list of choices of Didi.
(1<=len(s2)<=len(s1)<=1e6)
Output:
One line contains an integer indicating the maximum number of wining point.
忽略每行輸出的末尾多余空格
樣例輸入1
RRRRRRRRRLLL
RRRS
樣例輸出1
3
樣例輸入2
RSSPKKLLRKPS
RSRS
樣例輸出2
2
ACM-ICPC Asia Training League 寧夏理工學院
題解：
因為之前做過codeforces 528D. Fuzzy Search ，感覺就不難了，你要是不會這題可以先去做cf528d，有個詳細的題解：https://blog.csdn.net/u013368721/article/details/45565729
【FFT求字符串匹配的問題一般都是將模式串反轉，然后將其與主串進行卷積運算】
枚舉五種出拳方式，每種都做fft，最后掃一遍最大值即可求出最佳匹配出的贏的最大次數。（具體fft原理不懂orz，我就是套着原來板子寫的...）
#include<bits/stdc++.h>
#define CLR(a,b) memset((a),(b),sizeof((a)))
using namespace std; typedef long long ll; const int N = 1<<20; const double PI = acos(-1.0); int n, m; struct Complex { double x,y; Complex(double _x = 0.0,double _y = 0.0){ x = _x; y = _y; } Complex operator -(const Complex &b)const{ return Complex(x-b.x,y-b.y); } Complex operator +(const Complex &b)const{ return Complex(x+b.x,y+b.y); } Complex operator *(const Complex &b)const{ return Complex(x*b.x-y*b.y,x*b.y+y*b.x); } Complex operator * (const double &b)const{ return Complex(x * b,y * b); } Complex operator / (const double &b)const{ return Complex(x / b,y / b); } }; void change(Complex y[], int len) { int i, j, k; for(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].x /= len; } Complex a[N], b[N], c[N]; char s[N], t[N]; int sum[N]; int main() { int i, j, ans = 0, ma, nn; scanf("%s %s", s, t); n = strlen(s); m = strlen(t); reverse(t, t+m); ma = max(n, m); nn = 1; while(nn < 2 * ma) nn<<=1; CLR(c, 0); CLR(sum, 0); //R vs L S
CLR(a, 0); CLR(b, 0); for(i = 0; i < n; ++i) a[i].x = (s[i]=='L'||s[i]=='S'); for(i = 0; i < m; ++i) b[i].x = (t[i]=='R'); fft(a, nn, 1); fft(b, nn, 1); for(i = 0; i < nn ;++i) c[i] = a[i] * b[i]; fft(c, nn, -1); for(i = m-1; i < n; ++i) sum[i] += (int)(c[i].x+0.5); //P vs R K
CLR(a, 0); CLR(b, 0); for(i = 0; i < n; ++i) a[i].x = (s[i]=='R'||s[i]=='K'); for(i = 0; i < m; ++i) b[i].x = (t[i]=='P'); fft(a, nn, 1); fft(b, nn, 1); for(i = 0; i < nn ;++i) c[i] = a[i] * b[i]; fft(c, nn, -1); for(i = m-1; i < n; ++i) sum[i] += (int)(c[i].x+0.5); //S vs P L
CLR(a, 0); CLR(b, 0); for(i = 0; i < n; ++i) a[i].x = (s[i]=='P'||s[i]=='L'); for(i = 0; i < m; ++i) b[i].x = (t[i]=='S'); fft(a, nn, 1); fft(b, nn, 1); for(i = 0; i < nn ;++i) c[i] = a[i] * b[i]; fft(c, nn, -1); for(i = m-1; i < n; ++i) sum[i] += (int)(c[i].x+0.5); //L vs P K
CLR(a, 0); CLR(b, 0); for(i = 0; i < n; ++i) a[i].x = (s[i]=='P'||s[i]=='K'); for(i = 0; i < m; ++i) b[i].x = (t[i]=='L'); fft(a, nn, 1); fft(b, nn, 1); for(i = 0; i < nn ;++i) c[i] = a[i] * b[i]; fft(c, nn, -1); for(i = m-1; i < n; ++i) sum[i] += (int)(c[i].x+0.5); //K vs R S
CLR(a, 0); CLR(b, 0); for(i = 0; i < n; ++i) a[i].x = (s[i]=='R'||s[i]=='S'); for(i = 0; i < m; ++i) b[i].x = (t[i]=='K'); fft(a, nn, 1); fft(b, nn, 1); for(i = 0; i < nn ;++i) c[i] = a[i] * b[i]; fft(c, nn, -1); for(i = m-1; i < n; ++i) sum[i] += (int)(c[i].x+0.5); for(i = m-1; i < n; ++i) ans = max(ans, sum[i]); printf("%d\n", ans); return 0; }