Match & Catch CodeForces - 427D 后缀自动机水题

题意：

给出两个字符串a，b，求一个字符串，这个字符串是a和b的子串，

且只在a，b中出现一次，要求输出这个字符串的最小长度。

题解：

将a串放入后缀自动机中，然后记录一下每个节点对应的子串出现的次数

然后把b串取自动机中匹配

然后判断一下

 #include <set>
 #include <map>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <ctime>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <unordered_map>

 #define  pi    acos(-1.0)
 #define  eps   1e-9
 #define  fi    first
 #define  se    second
 #define  rtl   rt<<1
 #define  rtr   rt<<1|1
 #define  bug                printf("******\n")
 #define  mem(a, b)          memset(a,b,sizeof(a))
 #define  name2str(x)        #x
 #define  fuck(x)            cout<<#x" = "<<x<<endl
 #define  sfi(a)             scanf("%d", &a)
 #define  sffi(a, b)         scanf("%d %d", &a, &b)
 #define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 #define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  sfL(a)             scanf("%lld", &a)
 #define  sffL(a, b)         scanf("%lld %lld", &a, &b)
 #define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
 #define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
 #define  sfs(a)             scanf("%s", a)
 #define  sffs(a, b)         scanf("%s %s", a, b)
 #define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
 #define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
 #define  FIN                freopen("../in.txt","r",stdin)
 #define  gcd(a, b)          __gcd(a,b)
 #define  lowbit(x)          x&-x
 #define  IO                 iOS::sync_with_stdio(false)

 using namespace std;
 typedef long long LL;
 typedef unsigned long long ULL;
 const ULL seed = ;
 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 const int maxm = 8e6 + ;
 const int INF = 0x3f3f3f3f;
 const int mod = 1e9 + ;
 const int maxn = ;
 char s[maxn];
 int Q;

 struct Suffix_Automaton {
     int last, tot, nxt[maxn << ][], fail[maxn << ];//last是未加入此字符前最长的前缀（整个串）所属的节点的编号
     int len[maxn << ];// 最长子串的长度 (该节点子串数量 = len[x] - len[fa[x]])
     int sa[maxn << ], c[maxn << ];
     int sz[maxn << ];// 被后缀链接的个数，方便求节点字符串的个数
     LL num[maxn << ];// 该状态子串的数量
     LL maxx[maxn << ];// 长度为x的子串出现次数最多的子串的数目
     LL sum[maxn << ];// 该节点后面所形成的自字符串的总数
     LL subnum, sublen;// subnum表示不同字符串数目,sublen表示不同字符串总长度
     int X[maxn << ], Y[maxn << ]; // Y表示排名为x的节点，X表示该长度前面还有多少个
     int minn[maxn << ], mx[maxn << ];//minn[i]表示多个串在后缀自动机i节点最长公共子串，mx[i]表示单个串的最长公共子串
     void init() {
         tot = last = ;
         fail[] = len[] = ;
         for (int i = ; i < ; i++) nxt[][i] = ;
     }

     void extend(int c) {
         int u = ++tot, v = last;
         len[u] = len[v] + ;
         num[u] = ;
         for (; v && !nxt[v][c]; v = fail[v]) nxt[v][c] = u;
         if (!v) fail[u] = , sz[]++;
         else if (len[nxt[v][c]] == len[v] + ) fail[u] = nxt[v][c], sz[nxt[v][c]]++;
         else {
             int now = ++tot, cur = nxt[v][c];
             len[now] = len[v] + ;
             memcpy(nxt[now], nxt[cur], sizeof(nxt[cur]));
             fail[now] = fail[cur];
             fail[cur] = fail[u] = now;
             for (; v && nxt[v][c] == cur; v = fail[v]) nxt[v][c] = now;
             sz[now] += ;
         }
         last = u;
         //return len[last] - len[fail[last]];//多添加一个子串所产生不同子串的个数
     }

     void get_num() {// 每个节点子串出现的次数
         for (int i = ; i <= tot; i++) X[len[i]]++;
         for (int i = ; i <= tot; i++) X[i] += X[i - ];
         for (int i = ; i <= tot; i++) Y[X[len[i]]--] = i;
         for (int i = tot; i >= ; i--) num[fail[Y[i]]] += num[Y[i]];
     }

     void get_maxx(int n) {// 长度为x的子串出现次数最多的子串的数目
         get_num();
         for (int i = ; i <= tot; i++) maxx[len[i]] = max(maxx[len[i]], num[i]);
     }

     void get_sum() {// 该节点后面所形成的自字符串的总数
         get_num();
         for (int i = tot; i >= ; i--) {
             sum[Y[i]] = ;
             for (int j = ; j <= ; j++)
                 sum[Y[i]] += sum[nxt[Y[i]][j]];
         }
     }

     void get_subnum() {//本质不同的子串的个数
         subnum = ;
         for (int i = ; i <= tot; i++) subnum += len[i] - len[fail[i]];
     }

     void get_sublen() {//本质不同的子串的总长度
         sublen = ;
         for (int i = ; i <= tot; i++) sublen += 1LL * (len[i] + len[fail[i]] + ) * (len[i] - len[fail[i]]) / ;
     }

     void get_sa() { //获取sa数组
         for (int i = ; i <= tot; i++) c[len[i]]++;
         for (int i = ; i <= tot; i++) c[i] += c[i - ];
         for (int i = tot; i >= ; i--) sa[c[len[i]]--] = i;
     }

     int cntnum[maxn << ];

     void match() {//多个串的最长公共子串
         mem(cntnum, );
         int n = strlen(s), p = , ans = INF;
         for (int i = ; i < n; i++) {
             int c = s[i] - 'a';
             if (nxt[p][c]) p = nxt[p][c];
             else {
                 for (; p && !nxt[p][c]; p = fail[p]);
                 if (!p) p = ;
                 else p = nxt[p][c];
             }
             cntnum[p]++;
         }
         for (int i = tot; i; i--) cntnum[fail[Y[i]]] += cntnum[Y[i]];
         for (int i = ; i <= tot; i++)
             if (num[i] ==  && cntnum[i] == ) ans = min(ans, len[fail[i]] + );
         if (ans == INF) printf("-1\n");
         else printf("%d\n", ans);
     }

     void get_kth(int k) {//求出字典序第K的子串
         int pos = , cnt;
         string s = "";
         while (k) {
             for (int i = ; i <= ; i++) {
                 if (nxt[pos][i] && k) {
                     cnt = nxt[pos][i];
                     if (sum[cnt] < k) k -= sum[cnt];
                     else {
                         k--;
                         pos = cnt;
                         s += (char) (i + 'a');
                         break;
                     }
                 }
             }
         }
         cout << s << endl;
     }

 } sam;

 int main() {
 #ifndef ONLINE_JUDGE
     FIN;
 #endif
     sam.init();
     sfs(s + );
     int n = strlen(s + );
     for (int i = ; i <= n; i++) sam.extend((s[i] - 'a'));
     sfs(s);
     sam.get_num();
     sam.match();
 #ifndef ONLINE_JUDGE
     cout << "Totle Time : " << (double) clock() / CLOCKS_PER_SEC << "s" << endl;
 #endif
     return ;
 }