【LOJ】#2123. 「HEOI2015」最短不公共子串
题解
我们对于B串建出后缀自动机和序列自动机
对于问题1,枚举左端点然后跑后缀自动机,直到不能匹配作为这个左端点的答案
对于问题2,枚举左端点然后跑序列自动机,直到不能匹配
对于问题3,设f[i][j]表示第前i个字符匹配到后缀自动机上第j个点的最少步数,如果下一步走不了则更新答案
对于问题4,设f[i][j]表示前i个字符匹配到序列自动机上第j个点的最少步数,如果下一步走不了则更新答案
代码
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#define enter putchar('\n')
#define space putchar(' ')
//#define ivorysi
using namespace std;
typedef long long int64;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {putchar('-');x = -x;}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
const int MOD = 1000000007;
char A[2005],B[2005];
int LA,LB;
int f[2005][4005];
struct SuffixAutoMaton{
struct SAM_node {
SAM_node *par,*nxt[26];
int len;
SAM_node *find_next(char c) {
if(nxt[c - 'a']) return nxt[c - 'a'];
else return NULL;
}
}pool[4005],*tail = pool,*root,*last;
SuffixAutoMaton() {
root = last = tail++;
}
void build(int l,int e) {
SAM_node *nowp = tail++,*p;
nowp->len = l;
for(p = last ; p && !p->nxt[e]; p = p->par) {
p->nxt[e] = nowp;
}
if(!p) nowp->par = root;
else {
SAM_node *q = p->nxt[e];
if(q->len == p->len + 1) nowp->par = q;
else {
SAM_node *copyq = tail++;
*copyq = *q;
copyq->len = p->len + 1;
q->par = nowp->par = copyq;
for( ; p && p->nxt[e] == q ; p = p->par) {
p->nxt[e] = copyq;
}
}
}
last = nowp;
}
int calc1(int st) {
SAM_node *p = root;
int res = 0;
for(int i = st ; i <= LA ; ++i) {
p = p->find_next(A[i]);
if(!p) break;
++res;
}
if(res == LA - st + 1) return LA + 1;
return res + 1;
}
int calc3() {
int m = tail - pool;
for(int i = 0 ; i <= LA ; ++i) {
for(int j = 1 ; j <= m ; ++j) {
f[i][j] = 0x7fffffff;
}
}
f[0][1] = 0;
int res = LA + 1;
for(int i = 1 ; i <= LA ; ++i) {
for(int j = 1 ; j <= m ; ++j) {
if(f[i - 1][j] != 0x7fffffff) {
f[i][j] = min(f[i - 1][j],f[i][j]);
SAM_node *p = pool[j - 1].find_next(A[i]);
if(!p) res = min(res,f[i - 1][j] + 1);
else f[i][p - pool + 1] = min(f[i][p - pool + 1],f[i][j] + 1);
}
}
}
return res;
}
}SAM;
struct LineAutoMaton {
int ch[2005][26],head[30],next[2005],tot,rt;
LineAutoMaton() {
rt = tot = 1;
for(int i = 0 ; i < 26 ; ++i) head[i] = 1;
}
void build(char c) {
++tot;next[tot] = head[c - 'a'];
for(int i = 0 ; i < 26 ; ++i) {
for(int j = head[i] ; j ; j = next[j]) {
if(!ch[j][c - 'a']) ch[j][c - 'a'] = tot;
else break;
}
}
head[c - 'a'] = tot;
}
int calc2(int st) {
int p = rt,L = 0;
for(int i = st ; i <= LA ; ++i) {
if(ch[p][A[i] - 'a']) {
p = ch[p][A[i] - 'a'];
++L;
}
else break;
}
if(L == LA - st + 1) return LA + 1;
return L + 1;
}
int calc4() {
for(int i = 0 ;i <= LA ; ++i) {
for(int j = 1 ; j <= tot ; ++j) {
f[i][j] = 0x7fffffff;
}
}
f[0][1] = 0;
int res = LA + 1;
for(int i = 1 ; i <= LA ; ++i) {
for(int j = 1 ; j <= tot ; ++j) {
if(f[i - 1][j] != 0x7fffffff) {
f[i][j] = min(f[i - 1][j],f[i][j]);
int p = ch[j][A[i] - 'a'];
if(!p) res = min(res,f[i - 1][j] + 1);
else f[i][p] = min(f[i - 1][j] + 1,f[i][p]);
}
}
}
return res;
}
}LAM;
void Init() {
scanf("%s",A + 1);
scanf("%s",B + 1);
LA = strlen(A + 1);LB = strlen(B + 1);
for(int i = 1 ; i <= LB ; ++i) {
SAM.build(i,B[i] - 'a');
LAM.build(B[i]);
}
}
void Solve1() {
int ans = LA + 1;
for(int i = 1 ; i <= LA ; ++i) {
ans = min(ans,SAM.calc1(i));
}
if(ans > LA) puts("-1");
else {out(ans),enter;}
}
void Solve2() {
int ans = LA + 1;
for(int i = 1 ; i <= LA ; ++i) {
ans = min(ans,LAM.calc2(i));
}
if(ans > LA) puts("-1");
else out(ans),enter;
}
void Solve3() {
int ans = SAM.calc3();
if(ans > LA) puts("-1");
else out(ans),enter;
}
void Solve4() {
int ans = LAM.calc4();
if(ans > LA) puts("-1");
else out(ans),enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve1();
Solve2();
Solve3();
Solve4();
}