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http://acm.hdu.edu.cn/showproblem.php?pid=5769题意:在S串中找出X串出现的不同子串的数目? 其中1 <= |S| < $10^5$
官方题解: 处理出后缀数组中的sa[]数组和height[]数组。在不考虑包含字符X的情况下,不同子串的个数为
如果要求字符X,只需要记录距离sa[i]最近的字符X的位置(用nxt[sa[i]]表示)即可,个数
理解:后缀数组height[i]就是sa[i]与sa[i-1]的LCP,在后缀数组中求解全部的不同子串(之前只写过SAM处理所有不同子串..)还是比较好理解的,在一定要含有子串x时,需要先kmp求出所有匹配的位置,在处理到第i个后缀时,取max即可表示一定含有X,并且是不同的子串;
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<bits/stdc++.h> using namespace std; #define rep0(i,l,r) for(int i = (l);i < (r);i++) #define rep1(i,l,r) for(int i = (l);i <= (r);i++) #define rep_0(i,r,l) for(int i = (r);i > (l);i--) #define rep_1(i,r,l) for(int i = (r);i >= (l);i--) #define MS0(a) memset(a,0,sizeof(a)) #define MS1(a) memset(a,-1,sizeof(a)) #define MSi(a) memset(a,0x3f,sizeof(a)) #define inf 0x3f3f3f3f #define pb push_back #define MK make_pair #define A first #define B second #define eps 1e-8 #define zero(x) (((x)>0?(x):-(x))<eps) #define bitnum(a) __builtin_popcount(a) #define lowbit(x) (x&(-x)) #define clear0 (0xFFFFFFFE) #define mod 1000000007 #define K(x) ((x)*(x)) typedef pair<int,int> PII; typedef long long ll; typedef unsigned long long ull; template<typename T> void read1(T &m) { T x = 0,f = 1;char ch = getchar(); while(ch <'0' || ch >'9'){ if(ch == '-') f = -1;ch=getchar(); } while(ch >= '0' && ch <= '9'){ x = x*10 + ch - '0';ch = getchar(); } m = x*f; } template<typename T> void read2(T &a,T &b){read1(a);read1(b);} template<typename T> void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);} template<typename T> void out(T a) { if(a>9) out(a/10); putchar(a%10+'0'); } inline ll gcd(ll a,ll b){ return b == 0? a: gcd(b,a%b); } template<class T1, class T2> inline void gmax(T1& a, T2 b){ if(a < b) a = b;} template<class T1, class T2> inline void gmin(T1& a, T2 b){ if(a > b) a = b;} const int maxn = 100001; int sa[maxn], t[maxn],t2[maxn],c[maxn],w[maxn]; int cmp(int *r,int a,int b,int l) { return r[a] == r[b] && r[a+l] == r[b+l]; } void build_sa(char *r,int n,int m) { int i, j, p, *x = t, *y = t2; for(i = 0; i < m;i++) c[i] = 0; for(i = 0; i < n;i++) c[x[i] = r[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(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++) w[i] = x[y[i]]; for(i = 0; i < m; i++) c[i] = 0; for(i = 0; i < n; i++) c[w[i]]++; for(i = 0; i < m; i++) c[i] += c[i-1]; for(i = n-1; i >= 0; i--) sa[--c[w[i]]] = y[i]; swap(x,y); for(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++; if(p >= n) break; m = p; } } char S[maxn], X[maxn]; int height[maxn], rk[maxn]; void getHeight(int n) { for(int i = 1; i<= n; i++) rk[sa[i]] = i; for(int i = 0, j, k = 0; i < n; height[rk[i++]] = k) for(k? k--:0, j = sa[rk[i] - 1]; S[i+k] == S[j+k]; k++); } int f[maxn]; void getfail(char *p) { f[0] = f[1] = 0; int n = strlen(p); for(int i = 1;i < n;i++){ int j = f[i]; if(j && p[i] != p[j]) j = f[j]; f[i+1] = (p[i] == p[j] ?j+1:0);// i+1会递推到第n位 } } vector<int> vec; void Find(char *T, char *p) { vec.clear(); ll j = 0,n = strlen(T),m = strlen(p); for(int i = 0;i < n;i++){ while(j && T[i] != p[j]) j = f[j]; if(T[i] == p[j]) j++; if(j == m){ vec.pb(i); j = 0; i -= m-1; } } sort(vec.begin(), vec.end()); } int main() { //freopen("data.txt","r",stdin); //freopen("out.txt","w",stdout); int T, kase = 1; scanf("%d",&T); while(T--){ scanf("%s%s", X, S); int len = strlen(S), m = strlen(X); S[len] = '#';S[len+1] = 0; build_sa(S,len+1,'z'+1); getHeight(len); getfail(X); Find(S,X); vec.pb(len); ll ans = 0; rep1(i,1,len){ if(sa[i]+m-1 > len) continue; int nxt = lower_bound(vec.begin(), vec.end(), sa[i]+m-1) - vec.begin(); ans += len - max(vec[nxt],sa[i] + height[i]); } printf("Case #%d: %lld\n", kase++, ans); } return 0; }
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原文地址:http://www.cnblogs.com/hxer/p/5724311.html