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 <map>
3 #include <stack>
4 #include <queue>
5 #include <cmath>
6 #include <ctime>
7 #include <cstdio>
8 #include <string>
9 #include <vector>
10 #include <cstring>
11 #include <iostream>
12 #include <algorithm>
13 #include <unordered_map>
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" = "<<x<<endl
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 }
原文地址：https://www.cnblogs.com/qldabiaoge/p/11566941.html
时间： 2024-11-12 16:26:48