codeforces 954I （标算FFT,bitset水过）
这题题意非常之裸。定义一种变换是模糊两个字符，求s的每个与t长度相等的子串变成与t相等的的最小变换次数。
一眼就看出这是FFT，但是不会构造。
于是就看了一眼数据范围，125000，bitset可以水。（虽然好想好写，但是罚时爆炸，不过小号rating无关紧要）
思路就是和bitset字符串匹配的思路差不多。
复杂度O（m*(n-m)/计算机位数）
简写O(能过)
codeforces评测机是64位无疑。
然后大力卡常即可。
代码：
#pragma GCC target("avx") #pragma GCC optimize(3) #pragma GCC optimize("Ofast") #pragma GCC optimize("inline") #pragma GCC optimize("-fgcse") #pragma GCC optimize("-fgcse-lm") #pragma GCC optimize("-fipa-sra") #pragma GCC optimize("-ftree-pre") #pragma GCC optimize("-ftree-vrp") #pragma GCC optimize("-fpeephole2") #pragma GCC optimize("-ffast-math") #pragma GCC optimize("-fsched-spec") #pragma GCC optimize("unroll-loops") #pragma GCC optimize("-falign-jumps") #pragma GCC optimize("-falign-loops") #pragma GCC optimize("-falign-labels") #pragma GCC optimize("-fdevirtualize") #pragma GCC optimize("-fcaller-saves") #pragma GCC optimize("-fcrossjumping") #pragma GCC optimize("-fthread-jumps") #pragma GCC optimize("-funroll-loops") #pragma GCC optimize("-fwhole-program") #pragma GCC optimize("-freorder-blocks") #pragma GCC optimize("-fschedule-insns") #pragma GCC optimize("inline-functions") #pragma GCC optimize("-ftree-tail-merge") #pragma GCC optimize("-fschedule-insns2") #pragma GCC optimize("-fstrict-aliasing") #pragma GCC optimize("-fstrict-overflow") #pragma GCC optimize("-falign-functions") #pragma GCC optimize("-fcse-skip-blocks") #pragma GCC optimize("-fcse-follow-jumps") #pragma GCC optimize("-fsched-interblock") #pragma GCC optimize("-fpartial-inlining") #pragma GCC optimize("no-stack-protector") #pragma GCC optimize("-freorder-functions") #pragma GCC optimize("-findirect-inlining") #pragma GCC optimize("-fhoist-adjacent-loads") #pragma GCC optimize("-frerun-cse-after-loop") #pragma GCC optimize("inline-small-functions") #pragma GCC optimize("-finline-small-functions") #pragma GCC optimize("-ftree-switch-conversion") #pragma GCC optimize("-foptimize-sibling-calls") #pragma GCC optimize("-fexpensive-optimizations") #pragma GCC optimize("-funsafe-loop-optimizations") #pragma GCC optimize("inline-functions-called-once") #pragma GCC optimize("-fdelete-null-pointer-checks") #include<bits/stdc++.h> using namespace std; const int N1=20010,N2=40010,N3=80010,N4=125010,N5=60010; bitset<N1> ss[6],tt[6]; bitset<N2> sss[6],ttt[6]; bitset<N3> ssss[6],tttt[6]; bitset<N4> sssss[6],ttttt[6]; bitset<N5> ssssss[6],tttttt[6]; char s[N4],t[N4]; bool re[6][6],b[6]; int dfs(int x){ if (b[x]) return 0; b[x]=1; int ret=1; for (int i=0; i<6; ++i) if (re[x][i]||re[i][x]) ret+=dfs(i); return ret; } int main(){ scanf("%s%s",s,t); // freopen("re.out","w",stdout); int n=strlen(s),m=strlen(t); // n=120000,m=60000; // for (int i=0; i<n; ++i) s[i]='a'; // for (int i=0; i<m; ++i) t[i]='a'; if (m<=20000){ for (int i=0; i<m; ++i) ss[s[i]-'a'].set(i); for (int i=0; i<m; ++i) tt[t[i]-'a'].set(i); for (int u=0; u<=n-m; ++u){ int ret=0; for (int j=0; j<6; ++j) b[j]=0; for (int i=0; i<6; ++i) for (int j=0; j<6; ++j){ re[i][j]=0; if (j!=i&&(ss[i]&tt[j]).any()) re[i][j]=1; } for (int i=0; i<6; ++i) if (!b[i]) ret+=dfs(i)-1; printf("%d ",ret); for (int i=0; i<6; ++i) ss[i]>>=1; if (u<n-m) ss[s[u+m]-'a'].set(m-1); } } else if (m<=40000){ for (int i=0; i<m; ++i) sss[s[i]-'a'].set(i); for (int i=0; i<m; ++i) ttt[t[i]-'a'].set(i); for (int u=0; u<=n-m; ++u){ int ret=0; for (int j=0; j<6; ++j) b[j]=0; for (int i=0; i<6; ++i) for (int j=0; j<6; ++j){ re[i][j]=0; if (j!=i&&(sss[i]&ttt[j]).any()) re[i][j]=1; } for (int i=0; i<6; ++i) if (!b[i]) ret+=dfs(i)-1; printf("%d ",ret); for (int i=0; i<6; ++i) sss[i]>>=1; if (u<n-m) sss[s[u+m]-'a'].set(m-1); } } else if (m<=60010){ for (int i=0; i<m; ++i) ssssss[s[i]-'a'].set(i); for (int i=0; i<m; ++i) tttttt[t[i]-'a'].set(i); for (int u=0; u<=n-m; ++u){ int ret=0; for (int j=0; j<6; ++j) b[j]=0; for (int i=0; i<6; ++i) for (int j=0; j<6; ++j){ re[i][j]=0; if (j!=i&&(ssssss[i]&tttttt[j]).any()) re[i][j]=1; } for (int i=0; i<6; ++i) if (!b[i]) ret+=dfs(i)-1; printf("%d ",ret); for (int i=0; i<6; ++i) ssssss[i]>>=1; if (u<n-m) ssssss[s[u+m]-'a'].set(m-1); } } else if (m<=80010){ for (int i=0; i<m; ++i) ssss[s[i]-'a'].set(i); for (int i=0; i<m; ++i) tttt[t[i]-'a'].set(i); for (int u=0; u<=n-m; ++u){ int ret=0; for (int j=0; j<6; ++j) b[j]=0; for (int i=0; i<6; ++i) for (int j=0; j<6; ++j){ re[i][j]=0; if (j!=i&&(ssss[i]&tttt[j]).any()) re[i][j]=1; } for (int i=0; i<6; ++i) if (!b[i]) ret+=dfs(i)-1; printf("%d ",ret); for (int i=0; i<6; ++i) ssss[i]>>=1; if (u<n-m) ssss[s[u+m]-'a'].set(m-1); } } else{ for (int i=0; i<m; ++i) sssss[s[i]-'a'].set(i); for (int i=0; i<m; ++i) ttttt[t[i]-'a'].set(i); for (int u=0; u<=n-m; ++u){ int ret=0; for (int j=0; j<6; ++j) b[j]=0; for (int i=0; i<6; ++i) for (int j=0; j<6; ++j){ re[i][j]=0; if (j!=i&&(sssss[i]&ttttt[j]).any()) re[i][j]=1; } for (int i=0; i<6; ++i) if (!b[i]) ret+=dfs(i)-1; printf("%d ",ret); for (int i=0; i<6; ++i) sssss[i]>>=1; if (u<n-m) sssss[s[u+m]-'a'].set(m-1); } } }
bitset不支持可变长度，于是就分类讨论，反正bitset空间很小。实测比写的xiang的FFT快。
话说不是很懂Execution time Rank 1 的大佬的写法是什么？
