标签:ios using lse col set int mem get 题意
题意:
给你两串字符,要你找出在这两串字符中都出现过的最长子串
解析:
先用个分隔符将两个字符串连接起来,再用后缀数组求出height数组的值,找出一个height值最大并且i与i-1的sa值分别在两串字符中就好了
#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--) #define rd(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; } } char s1[maxn], s2[maxn]; int main() { while(~scanf("%s%s", s1, s2)){ n = 0; int len1 = strlen(s1); rep(i, 0, len1) s[n++] = s1[i] - ‘a‘ + 1; s[n++] = 28; int len2 = strlen(s2); rep(i, 0, len2) s[n++] = s2[i] - ‘a‘ + 1; s[n++] = 0; get_sa(30); int maxx = -INF; rep(i, 1, n) { if(height[i] > maxx && sa[i] < len1 && sa[i-1] > len1 && sa[i] >= 0) maxx = height[i]; else if(height[i] > maxx && sa[i-1] < len1 && sa[i] > len1 && sa[i-1] >= 0) maxx = height[i]; } cout<< maxx <<endl; } return 0; }
标签:ios using lse col set int mem get 题意
原文地址:https://www.cnblogs.com/WTSRUVF/p/9495416.html