spoj1811 Longest Common Substring(LCS)最长公共子串
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spoj1811LCS
问两个字符串最长公共子串。
做法很简单。匹配成功,则tl++,失败,从父指针回退,tl=t[now].len。
指向一个状态,这个状态的接受串s[x..x+i]是与当前状态的接受串后缀s[j-i..j]匹配是最长的一个。
这里是不是发现了一个和KMP很像的性质?
KMP在失配时通过next数组回退,那么这个回退到的位置i是s[0..i]与当前串的后缀s[j-i..j]匹配最长的一个。
所以。
利用后缀自动机可以求解一个串的子串(s[x..])与另一个串的子串的最长匹配长度。
KMP可以求解一个串(s[0..])与另一个串的子串的最长匹配长度。
//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef pair<long long int,long long int> ii;typedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=3e5+10;const int maxx=600005;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;int root,last;int tots; //总结点int l; //字符串长度struct sam_node{ int fa,son[26]; int len; void init(int _len) { len = _len; fa = -1; memset(son,-1,sizeof(son)); }}t[maxn*2];void sam_init(){ tots = 0; root = last = 0; t[tots].init(0);}void extend(int w){ int p=last; int np=++tots;t[tots].init(t[p].len+1); int q,nq; while(p!=-1&&t[p].son[w]==-1){t[p].son[w]=np;p=t[p].fa;} if (p==-1) t[np].fa=root; else { q=t[p].son[w]; if (t[p].len+1==t[q].len){t[np].fa=q;} else { nq=++tots;t[nq].init(0); t[nq]=t[q]; t[nq].len=t[p].len+1; t[q].fa=nq;t[np].fa=nq; while(p!=-1&&t[p].son[w]==q){t[p].son[w]=nq;p=t[p].fa;} } } last=np;}int w[maxn], r[maxn*2];void topsort(){ for(int i = 0; i <= l; ++i) w[i] = 0; for(int i = 1; i <= tots; ++i) w[t[i].len]++; for(int i = 1; i <= l; ++i) w[i] += w[i-1]; for(int i = tots; i >= 1; --i) r[w[t[i].len]--] = i; r[0] = 0;}//int dp[maxn*2];char s[maxn];char f[maxn];int work(int l2){ int i,now=root,ind,tl=0; int ret=0; for(i=0;i<l2;++i) { ind=f[i]-'a'; while(now!=-1&&t[now].son[ind]==-1) { now=t[now].fa; if (now!=-1) tl=t[now].len; } if (now==-1) {now=root;tl=0;} else { now=t[now].son[ind]; tl++; ret=max(ret,tl); } } return ret;}int main(){ int l1,l2,ans=0; scanf("%s",s); scanf("%s",f); l1=strlen(s); l2=strlen(f); sam_init(); FOr(0,l1,i) extend(s[i]-'a'); ans=work(l2); print(ans);}
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