hdu 5769 Substring 后缀数组 + KMP

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

hdu 5769 Substring 后缀数组 + KMP

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

hdu 5769 Substring 后缀数组 + KMP

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 = ,f = ;char ch = getchar();
while(ch <'' || ch >''){ if(ch == '-') f = -;ch=getchar(); }
while(ch >= '' && ch <= ''){ x = x* + ch - '';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>) out(a/);
putchar(a%+'');
}
inline ll gcd(ll a,ll b){ return b == ? 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 = ;
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 = ; i < m;i++) c[i] = ;
for(i = ; i < n;i++) c[x[i] = r[i]]++;
for(i = ; i < m;i++) c[i] += c[i-];
for(i = n-; i >= ; i--) sa[--c[x[i]]] = i; for(j = , p = ; p < n; j <<= , m = p){
for(p = , i = n - j; i < n;i++) y[p++] = i;
for(i = ; i < n; i++) if(sa[i] >= j) y[p++] = sa[i]-j;
for(i = ; i < n; i++) w[i] = x[y[i]];
for(i = ; i < m; i++) c[i] = ;
for(i = ; i < n; i++) c[w[i]]++;
for(i = ; i < m; i++) c[i] += c[i-];
for(i = n-; i >= ; i--) sa[--c[w[i]]] = y[i];
swap(x,y);
for(p = , x[sa[]] = , i = ; i < n; i++)
x[sa[i]] = cmp(y, sa[i-], sa[i], j)? p-: p++;
if(p >= n) break;
m = p;
}
}
char S[maxn], X[maxn];
int height[maxn], rk[maxn];
void getHeight(int n)
{
for(int i = ; i<= n; i++) rk[sa[i]] = i;
for(int i = , j, k = ; i < n; height[rk[i++]] = k)
for(k? k--:, j = sa[rk[i] - ]; S[i+k] == S[j+k]; k++);
} int f[maxn];
void getfail(char *p)
{
f[] = f[] = ;
int n = strlen(p);
for(int i = ;i < n;i++){
int j = f[i];
if(j && p[i] != p[j]) j = f[j];
f[i+] = (p[i] == p[j] ?j+:);// i+1会递推到第n位
}
}
vector<int> vec;
void Find(char *T, char *p)
{
vec.clear();
ll j = ,n = strlen(T),m = strlen(p);
for(int i = ;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 = ;
i -= m-;
}
}
sort(vec.begin(), vec.end());
} int main()
{
//freopen("data.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T, kase = ;
scanf("%d",&T);
while(T--){
scanf("%s%s", X, S);
int len = strlen(S), m = strlen(X);
S[len] = '#';S[len+] = ;
build_sa(S,len+,'z'+);
getHeight(len); getfail(X);
Find(S,X);
vec.pb(len);
ll ans = ;
rep1(i,,len){
if(sa[i]+m- > len) continue;
int nxt = lower_bound(vec.begin(), vec.end(), sa[i]+m-) - vec.begin();
ans += len - max(vec[nxt],sa[i] + height[i]);
}
printf("Case #%d: %lld\n", kase++, ans);
}
return ;
}
上一篇:将本地文件通过Xftp上传到服务器报错:permision is not allowed


下一篇:数据中心业务连续性和灾难恢复:计划和测试间的差距