CodeForces

题意:求母串中可以匹配模式串的子串的个数,但是每一位i的字符可以左右偏移k个位置.
分析:类似于 UVALive -4671. 用FFT求出每个字符成功匹配的个数.因为字符可以偏移k个单位,先用尺取法处理出每个位置能够取到的字符.设模式串长度为m.
令(C(m-1+k) = sum_{i=0}^{m-1}A_{i+k}*B(m-i-1)).
反转模式串B, 对每个字符c,若该位上能够取到c,则多项式该位取1,否则为0,FFT求卷积.并记录[m-1,n-1]每个位置4次计算的系数(C)之和.
若系数之和=m,表示母串中以位置i结尾,长度为m的字串与B相匹配.

#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 的数量
    //f g和 的系数
    // 卷积结果
    // 大数乘积
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(){             //求多项式
    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));
}

const int len = 2e5+10;

char s1[len],s2[len];
int cnt[4];
int have[MAXN][4];
map<char,int> id;
LL ans[MAXN];

void debug(){
    for(int i=0;i<4;++i){
        for(int j=0;j<4;++j){
            cout<<have[i][j]<<" ";
        }
        cout<<endl;
    }
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    int n,m,k;
    id['A'] = 0, id['C'] = 1, id['G'] =2, id['T'] = 3;
    scanf("%d %d %d",&n,&m,&k );
    scanf("%s",s1);
    scanf("%s",s2);
    int L=0,R=-1;
    for(int i=0;i<n;++i){
        while(L<i-k) cnt[id[s1[L++]]]--;         //退
        while(R<n-1 && R<i+k) cnt[id[s1[++R]]]++;  //增
        for(int j=0;j<4;++j){
            if(cnt[j]) have[i][j] = 1;
        }
    }
    for(int k=0;k<4;++k){
        N = n, M = m;
        for(int i=0;i<N;++i){
            if(have[i][k]) f[i] = 1;
            else f[i] = 0;
        }
        for(int i=0;i<M;++i){
            if(id[s2[m-i-1]]==k) g[i] = 1;
            else g[i] = 0;
        }
        getConv();
        int sz = conv.size();
        for(int i=m-1;i<sz;++i){
            ans[i]+= conv[i];
        }
    }
    int res=0;
    for(int i=m-1;i<n;++i){
        if(ans[i]==m){
            res++;
        }
    }
    printf("%d
",res);
    return 0;
}