Common Substrings POJ - 3415(長度不小於k的公共子串的個數)
阿新 • • 發佈：2018-08-18
mat oid continue return src substr alt 技術 pen
題意：
給定兩個字符串A 和 B， 求長度不小於 k 的公共子串的個數（可以相同）
分兩部分求和sa[i-1] > len1 sa[i] < len1 和 sa[i-1] < len1 sa[i] > len1
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include<vector> #include <stack> #include <queue> #include <algorithm> #include <cmath> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #define lep(i, a, n) for(int i=n; i>a; i--) #definerd(a) scanf("%d", &a) #define rlld(a) scanf("%lld", &a) #define rc(a) scanf("%c", &a) #define rs(a) scanf("%s", a) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 200010, INF = 0x7fffffff; int s[maxn]; int sa[maxn], t[maxn], t2[maxn], c[maxn], n; int ran[maxn], height[maxn]; void get_sa(int m) { int i, *x = t, *y = t2; for(i = 0; i < m; i++) c[i] = 0; for(i = 0; i < n; i++) c[x[i] = s[i]]++; for(i = 1; i < m; i++) c[i] += c[i-1]; for(i = n-1; i >= 0; i--) sa[--c[x[i]]] = i; for(int k = 1; k <= n; k <<= 1) { int p = 0; for(i = n-k; i < n; i++) y[p++] = i; for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k; for(i = 0; i < m; i++) c[i] = 0; for(i = 0; i < n; i++) c[x[y[i]]]++; for(i = 0; i< m; i++) c[i] += c[i-1]; for(i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i]; swap(x, y); p = 1; x[sa[0]] = 0; for(i = 1; i < n; i++) x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++; if(p >= n) break; m = p; } int k = 0; for(i = 0; i < n; i++) ran[sa[i]] = i; for(i = 0; i < n; i++) { if(k) k--; int j = sa[ran[i]-1]; while(s[i+k] == s[j+k]) k++; height[ran[i]] = k; } } int k, top, num; LL sum, ans; char s1[maxn], s2[maxn]; int stac[maxn], cnt[maxn]; int main() { while(~rd(k) && k) { top = sum = num = ans = n = 0; rs(s1); rs(s2); int len1 = strlen(s1); int len2 = strlen(s2); rep(i, 0, len1) s[n++] = s1[i]; s[n++] = ‘#‘; rep(i, 0, len2) s[n++] = s2[i]; s[n++] = 0; get_sa(200); rep(i, 1, n) { if(height[i] < k) { sum = top = 0; continue; } int num = 0; while(top && height[i] < stac[top]) //維持單調遞增棧 可能當前sa[i-1] < len1 但height是連續的 所以短板效應替換棧中元素 { //而它自己如果sa[i-1] < len1 那麽下面的 num是不加1的 即自己不算在內 sum -= (LL)(stac[top] - k + 1) * cnt[top]; sum += (LL)(height[i] - k + 1) * cnt[top]; num += cnt[top]; top--; } stac[++top] = height[i]; if(sa[i-1] > len1) //掃描B串 { sum += (LL)(height[i] - k + 1); cnt[top] = num + 1; } else cnt[top] = num; if(sa[i] < len1) ans += sum; } rep(i, 1, n) { if(height[i] < k) { sum = top = 0; continue; } int num = 0; while(top && height[i] < stac[top]) { sum -= (LL)(stac[top] - k + 1) * cnt[top]; sum += (LL)(height[i] - k + 1) * cnt[top]; num += cnt[top]; top--; } stac[++top] = height[i]; if(sa[i-1] < len1) //掃描A串 { sum += (LL)(height[i] - k + 1); cnt[top] = num + 1; } else cnt[top] = num; if(sa[i] > len1) ans += sum; } printf("%lld\n", ans); } return 0; }
Common Substrings POJ - 3415(長度不小於k的公共子串的個數)