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

时间:2019-09-22 12:50:24      阅读:93      评论:0      收藏:0      [点我收藏+]

标签:后缀   emc   最长公共子串   scanf   kth   bnu   for   ORC   mod   

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

 

技术图片
  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 }
View Code

 

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

标签:后缀   emc   最长公共子串   scanf   kth   bnu   for   ORC   mod   

原文地址:https://www.cnblogs.com/qldabiaoge/p/11566941.html

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