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
1 #include <set> 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include <string> 9 #include 10 #include 11 #include 12 #include 13 #include 14 15 #define pi acos(-1.0) 16 #define eps 1e-9 17 #define fi first 18 #define se second 19 #define rtl rt<<1 20 #define rtr rt<<1|1 21 #define bug printf("******\n") 22 #define mem(a, b) memset(a,b,sizeof(a)) 23 #define name2str(x) #x 24 #define fuck(x) cout<<#x" = "< 25 #define sfi(a) scanf("%d", &a) 26 #define sffi(a, b) scanf("%d %d", &a, &b) 27 #define sfffi(a, b, c) scanf("%d %d %d", &a, &b, &c) 28 #define sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d) 29 #define sfL(a) scanf("%lld", &a) 30 #define sffL(a, b) scanf("%lld %lld", &a, &b) 31 #define sfffL(a, b, c) scanf("%lld %lld %lld", &a, &b, &c) 32 #define sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d) 33 #define sfs(a) scanf("%s", a) 34 #define sffs(a, b) scanf("%s %s", a, b) 35 #define sfffs(a, b, c) scanf("%s %s %s", a, b, c) 36 #define sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d) 37 #define FIN freopen("../in.txt","r",stdin) 38 #define gcd(a, b) __gcd(a,b) 39 #define lowbit(x) x&-x 40 #define IO iOS::sync_with_stdio(false) 41 42 43 using namespace std; 44 typedef long long LL; 45 typedef unsigned long long ULL; 46 const ULL seed = 13331; 47 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL; 48 const int maxm = 8e6 + 10; 49 const int INF = 0x3f3f3f3f; 50 const int mod = 1e9 + 7; 51 const int maxn = 250007; 52 char s[maxn]; 53 int Q; 54 55 struct Suffix_Automaton { 56 int last, tot, nxt[maxn << 1][26], fail[maxn << 1];//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 57 int len[maxn << 1];// 最长子串的长度 (该节点子串数量 = len[x] - len[fa[x]]) 58 int sa[maxn << 1], c[maxn << 1]; 59 int sz[maxn << 1];// 被后缀链接的个数，方便求节点字符串的个数 60 LL num[maxn << 1];// 该状态子串的数量 61 LL maxx[maxn << 1];// 长度为x的子串出现次数最多的子串的数目 62 LL sum[maxn << 1];// 该节点后面所形成的自字符串的总数 63 LL subnum, sublen;// subnum表示不同字符串数目,sublen表示不同字符串总长度 64 int X[maxn << 1], Y[maxn << 1]; // Y表示排名为x的节点，X表示该长度前面还有多少个 65 int minn[maxn << 1], mx[maxn << 1];//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 66 void init() { 67 tot = last = 1; 68 fail[1] = len[1] = 0; 69 for (int i = 0; i < 26; i++) nxt[1][i] = 0; 70 } 71 72 void extend(int c) { 73 int u = ++tot, v = last; 74 len[u] = len[v] + 1; 75 num[u] = 1; 76 for (; v && !nxt[v][c]; v = fail[v]) nxt[v][c] = u; 77 if (!v) fail[u] = 1, sz[1]++; 78 else if (len[nxt[v][c]] == len[v] + 1) fail[u] = nxt[v][c], sz[nxt[v][c]]++; 79 else { 80 int now = ++tot, cur = nxt[v][c]; 81 len[now] = len[v] + 1; 82 memcpy(nxt[now], nxt[cur], sizeof(nxt[cur])); 83 fail[now] = fail[cur]; 84 fail[cur] = fail[u] = now; 85 for (; v && nxt[v][c] == cur; v = fail[v]) nxt[v][c] = now; 86 sz[now] += 2; 87 } 88 last = u; 89 //return len[last] - len[fail[last]];//多添加一个子串所产生不同子串的个数 90 } 91 92 void get_num() {// 每个节点子串出现的次数 93 for (int i = 1; i <= tot; i++) X[len[i]]++; 94 for (int i = 1; i <= tot; i++) X[i] += X[i - 1]; 95 for (int i = 1; i <= tot; i++) Y[X[len[i]]--] = i; 96 for (int i = tot; i >= 1; i--) num[fail[Y[i]]] += num[Y[i]]; 97 } 98 99 void get_maxx(int n) {// 长度为x的子串出现次数最多的子串的数目 100 get_num(); 101 for (int i = 1; i <= tot; i++) maxx[len[i]] = max(maxx[len[i]], num[i]); 102 } 103 104 void get_sum() {// 该节点后面所形成的自字符串的总数 105 get_num(); 106 for (int i = tot; i >= 1; i--) { 107 sum[Y[i]] = 1; 108 for (int j = 0; j <= 25; j++) 109 sum[Y[i]] += sum[nxt[Y[i]][j]]; 110 } 111 } 112 113 void get_subnum() {//本质不同的子串的个数 114 subnum = 0; 115 for (int i = 1; i <= tot; i++) subnum += len[i] - len[fail[i]]; 116 } 117 118 void get_sublen() {//本质不同的子串的总长度 119 sublen = 0; 120 for (int i = 1; i <= tot; i++) sublen += 1LL * (len[i] + len[fail[i]] + 1) * (len[i] - len[fail[i]]) / 2; 121 } 122 123 void get_sa() { //获取sa数组 124 for (int i = 1; i <= tot; i++) c[len[i]]++; 125 for (int i = 1; i <= tot; i++) c[i] += c[i - 1]; 126 for (int i = tot; i >= 1; i--) sa[c[len[i]]--] = i; 127 } 128 129 int cntnum[maxn << 1]; 130 131 void match() {//多个串的最长公共子串 132 mem(cntnum, 0); 133 int n = strlen(s), p = 1, ans = INF; 134 for (int i = 0; i < n; i++) { 135 int c = s[i] - 'a'; 136 if (nxt[p][c]) p = nxt[p][c]; 137 else { 138 for (; p && !nxt[p][c]; p = fail[p]); 139 if (!p) p = 1; 140 else p = nxt[p][c]; 141 } 142 cntnum[p]++; 143 } 144 for (int i = tot; i; i--) cntnum[fail[Y[i]]] += cntnum[Y[i]]; 145 for (int i = 2; i <= tot; i++) 146 if (num[i] == 1 && cntnum[i] == 1) ans = min(ans, len[fail[i]] + 1); 147 if (ans == INF) printf("-1\n"); 148 else printf("%d\n", ans); 149 } 150 151 void get_kth(int k) {//求出字典序第K的子串 152 int pos = 1, cnt; 153 string s = ""; 154 while (k) { 155 for (int i = 0; i <= 25; i++) { 156 if (nxt[pos][i] && k) { 157 cnt = nxt[pos][i]; 158 if (sum[cnt] < k) k -= sum[cnt]; 159 else { 160 k--; 161 pos = cnt; 162 s += (char) (i + 'a'); 163 break; 164 } 165 } 166 } 167 } 168 cout << s << endl; 169 } 170 171 } sam; 172 173 int main() { 174 #ifndef ONLINE_JUDGE 175 FIN; 176 #endif 177 sam.init(); 178 sfs(s + 1); 179 int n = strlen(s + 1); 180 for (int i = 1; i <= n; i++) sam.extend((s[i] - 'a')); 181 sfs(s); 182 sam.get_num(); 183 sam.match(); 184 #ifndef ONLINE_JUDGE 185 cout << "Totle Time : " << (double) clock() / CLOCKS_PER_SEC << "s" << endl; 186 #endif 187 return 0; 188 }
