NC20453. [TJOI2017]DNA
描述
输入描述
第一行有一个数T,表示有几组数据每组数据第一行一个长度不超过10^5的碱基序列S0每组数据第二行一个长度不超过10^5的吃藕基因序列S
输出描述
共T行,第i行表示第i组数据中,在S0中有多少个与S等长的连续子串可能是表现吃藕性状的碱基序列
示例1
输入:
1 ATCGCCCTA CTTCA
输出:
2
C++14(g++5.4) 解法, 执行用时: 466ms, 内存消耗: 21368K, 提交时间: 2020-09-03 18:36:30
#include<iostream> #include<algorithm> using namespace std; #define re register const int max_n = 2e5 + 100; int ranks[max_n], SA[max_n], height[max_n]; int wa[max_n], wb[max_n], wvarr[max_n], wsarr[max_n]; inline int cmp(int* r, int a, int b, int l) { return r[a] == r[b] && r[a + l] == r[b + l]; } inline void get_sa(int* r, int* sa, int n, int m) { ++n; re int i, j, p, * x = wa, * y = wb, * t; for (i = 0; i < m; ++i) wsarr[i] = 0; for (i = 0; i < n; ++i) wsarr[x[i] = r[i]]++; for (i = 1; i < m; ++i) wsarr[i] += wsarr[i - 1]; for (i = n - 1; i >= 0; --i) sa[--wsarr[x[i]]] = i; for (j = 1, p = 1; p < n; j <<= 1, m = p) { for (p = 0, i = n - j; i < n; ++i) y[p++] = i; for (i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j; for (i = 0; i < n; ++i) wvarr[i] = x[y[i]]; for (i = 0; i < m; ++i) wsarr[i] = 0; for (i = 0; i < n; ++i) wsarr[wvarr[i]]++; for (i = 1; i < m; ++i) wsarr[i] += wsarr[i - 1]; for (i = n - 1; i >= 0; --i) sa[--wsarr[wvarr[i]]] = y[i]; for (t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; ++i) x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++; } } void get_height(int* r, int* sa, int n) { re int i, j, k = 0; for (i = 0; i <= n; ++i) ranks[sa[i]] = i; for (i = 0; i < n; height[ranks[i++]] = k) for (k ? k-- : 0, j = sa[ranks[i] - 1]; r[i + k] == r[j + k]; k++); return; } int st[max_n][32]; void initSt(int a[], int n) { for (int i = 0;i <= n;++i)st[i][0] = height[i]; int mxk = (int)log2(n + 1); for (int k = 1;k <= mxk;++k) { for (int i = 0;i <= n;++i) { if (i + (1 << k) - 1 > n)break; st[i][k] = min(st[i][k - 1], st[i + (1 << (k - 1))][k - 1]); } } } int que(int l, int r) { l = ranks[l];r = ranks[r]; if (l > r)swap(l, r); ++l; int len = log2(r - l + 1); return min(st[l][len], st[r - (1 << len) + 1][len]); } int a[max_n]; string s1, s2, s3; void init(int n) { fill(a, a + n + 3, 0); fill(ranks, ranks + n + 3, 0); fill(SA, SA + n + 3, 0); fill(height, height + n + 3, 0); fill(wa, wa + n + 3, 0); fill(wb, wb + n + 3, 0); fill(wsarr, wsarr + n + 3, 0); fill(wvarr, wvarr + n + 3, 0); } int ToInt(char ch) { if (ch == 'A')return 1; else if (ch == 'T')return 2; else if (ch == 'C')return 3; else if (ch == 'G')return 4; else return 5; } int main() { ios::sync_with_stdio(0); int T;cin >> T; while (T--) { cin >> s1;cin >> s2; s3 = s1 + s2; int n = s3.size(); init(n); for (re int i = 0;i < n;++i)a[i] = ToInt(s3[i]); get_sa(a, SA, n, 7); get_height(a, SA, n); initSt(a, n); int ans = 0; for (re int i = s2.size();i <= s1.size();++i) { int left = i - s2.size(); int cur = left; for (int k = 0;k <= 3;++k) { cur += que(cur, s1.size() + cur - left); if (k < 3)++cur; if (cur >= i) { ++ans; break; } } }cout << ans << endl; } }
C++(clang++11) 解法, 执行用时: 82ms, 内存消耗: 3836K, 提交时间: 2021-02-13 15:33:10
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define N 100050 #define LL unsigned long long int T,l1,l2; char s1[N],s2[N]; LL mi[N],p=19260817,h1[N],h2[N]; bool check(int x) { int i=x,j=1; int l=1,r=l2+1,mid; for(int k=1;k<=3;++k) { while(l<r) { mid=(l+r)>>1; if(h1[i+mid-1]-h1[i-1]*mi[mid]==h2[j+mid-1]-h2[j-1]*mi[mid]) l=mid+1; else r=mid; } i=i+l; j=j+l; l=1; r=l2-j+2; if(j>l2) return 1; mid=l2-j+1; if(h1[i+mid-1]-h1[i-1]*mi[mid]==h2[j+mid-1]-h2[j-1]*mi[mid]) return 1; } return 0; } int main() { scanf("%d",&T); while(T--) { int i,ans=0; scanf("%s%s",s1+1,s2+1); l1=strlen(s1+1); l2=strlen(s2+1); if(l1<l2) { puts("0"); continue; } mi[0]=1; for(i=1;i<=l1;++i) { mi[i]=mi[i-1]*p; h1[i]=h1[i-1]*p+s1[i]; } for(i=1;i<=l2;++i) h2[i]=h2[i-1]*p+s2[i]; for(i=1;i<=l1-l2+1;++i) { if(check(i)) ans++; } printf("%d\n",ans); } }