UVALive - 4671 K-neighbor substrings (FFT+哈希)
阿新 • • 發佈:2018-10-02
pre eof ++i 字符串 swa subst can pac sin ,則蓋子串符合條件,但本題需要求不同的子串,所以將母串哈希掉,將符合要求的哈希值存在集合中去重.
題意:海明距離的定義:兩個相同長度的字符串中不同的字符數.現給出母串A和模式串B,求A中有多少與B海明距離<=k的不同子串
分析:將字符a視作1,b視作0.則A與B中都是a的位置乘積是1.現將B逆置,並設B的長度為n,令\(C(n+k-1)= \sum_{i=0}^{n-1}A_{i+k}*B_{n-i-1}\),表示母串A中從位置k開始,長度為n的子串與B中字符都是‘a‘的位置的數目,可以通過FFT運算得到.再對字符‘b‘做一次同樣的運算,\(ans[i]\)統計母串A中以i結尾的子串與B相同字符的個數.
設A的長度為m,則一共有\(m-n+1\)個子串,若\(n-ans[i] \leq k\)
#include <bits/stdc++.h> using namespace std; typedef long long LL; const int MAXN = 4e5 + 10; const double PI = acos(-1.0); struct Complex{ double x, y; inline Complex operator+(const Complex b) const { return (Complex){x +b.x,y + b.y}; } inline Complex operator-(const Complex b) const { return (Complex){x -b.x,y - b.y}; } inline Complex operator*(const Complex b) const { return (Complex){x *b.x -y * b.y,x * b.y + y * b.x}; } } va[MAXN * 2 + MAXN / 2], vb[MAXN * 2 + MAXN / 2]; int lenth = 1, rev[MAXN * 2 + MAXN / 2]; int N, M; // f 和 g 的數量 //f g和 的系數 // 卷積結果 // 大數乘積 int f[MAXN],g[MAXN]; vector<LL> conv; vector<LL> multi; //f g void init() { int tim = 0; lenth = 1; conv.clear(), multi.clear(); memset(va, 0, sizeof va); memset(vb, 0, sizeof vb); while (lenth <= N + M - 2) lenth <<= 1, tim++; for (int i = 0; i < lenth; i++) rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (tim - 1)); } void FFT(Complex *A, const int fla) { for (int i = 0; i < lenth; i++){ if (i < rev[i]){ swap(A[i], A[rev[i]]); } } for (int i = 1; i < lenth; i <<= 1){ const Complex w = (Complex){cos(PI / i), fla * sin(PI / i)}; for (int j = 0; j < lenth; j += (i << 1)){ Complex K = (Complex){1, 0}; for (int k = 0; k < i; k++, K = K * w){ const Complex x = A[j + k], y = K * A[j + k + i]; A[j + k] = x + y; A[j + k + i] = x - y; } } } } void getConv(){ //求多項式 init(); for (int i = 0; i < N; i++) va[i].x = f[i]; for (int i = 0; i < M; i++) vb[i].x = g[i]; FFT(va, 1), FFT(vb, 1); for (int i = 0; i < lenth; i++) va[i] = va[i] * vb[i]; FFT(va, -1); for (int i = 0; i <= N + M - 2; i++) conv.push_back((LL)(va[i].x / lenth + 0.5)); } char s1[100005],s2[100005]; LL res[MAXN]; const int seed = 3; LL dig[MAXN],Hash[MAXN]; set<LL> dp; void pre(){ dig[0] =1; for(int i=1;i<=100005;++i){ dig[i] = dig[i-1]*seed; } } LL getHash(int L ,int R){ if(L==0) return Hash[R]; return Hash[R] - Hash[L-1]*dig[R-L+1]; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); freopen("out.txt","w",stdout); #endif pre(); int k,cas=1; while(scanf("%d",&k)==1){ if(k==-1) break; memset(res,0,sizeof(res)); dp.clear(); scanf("%s",s1); scanf("%s",s2); int len1 = strlen(s1), len2 = strlen(s2); N = len1, M = len2; for(int i=0;i<len1;++i){ if(s1[i]==‘a‘) f[i] = 1; else f[i] = 0; } for(int i=0;i<len2;++i){ if(s2[len2-i-1]==‘a‘) g[i] = 1; else g[i] = 0; } getConv(); int sz =conv.size(); for(int i=len2-1;i<sz;++i){ res[i] += conv[i]; } for(int i=0;i<len1;++i){ if(s1[i]==‘b‘) f[i] = 1; else f[i] = 0; } for(int i=0;i<len2;++i){ if(s2[len2-i-1]==‘b‘) g[i] = 1; else g[i] = 0; } getConv(); sz =conv.size(); for(int i=len2-1;i<sz;++i){ res[i] += conv[i]; } //Hash Hash[0] = s1[0]-‘a‘+1; for(int i=1;i<len1;++i){ Hash[i] = Hash[i-1]*seed + s1[i]-‘a‘+1; } for(int i=len2-1;i<len1;++i){ LL now = getHash(i-len2+1,i); if(len2-res[i]<=k){ dp.insert(now); } } printf("Case %d: %d\n",cas++,(int)dp.size()); } return 0; }
UVALive - 4671 K-neighbor substrings (FFT+哈希)