长度为x的本质不同的串的出现次数 SPOJ - NSUBSTR 后缀自动机简单应用

题意:

 

长度为x的本质不同的串的出现次数 

 

题解:

 

先处理出每一个节点所对应的子串出现的次数

然后取max就好了

 

  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 
 53 struct Suffix_Automaton {
 54     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
 55     int len[maxn << 1];// 最长子串的长度 (该节点子串数量 = len[x] - len[fa[x]])
 56     int sa[maxn << 1], c[maxn << 1];
 57     int sz[maxn << 1];// 被后缀链接的个数,方便求节点字符串的个数
 58     LL num[maxn << 1];// 该状态子串的数量
 59     LL maxx[maxn << 1];// 长度为x的子串出现次数最多的子串的数目
 60     LL sum[maxn << 1];// 该节点后面所形成的自字符串的总数
 61     LL subnum, sublen;// subnum表示不同字符串数目,sublen表示不同字符串总长度
 62     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
 63     void init() {
 64         tot = last = 1;
 65         fail[1] = len[1] = 0;
 66         for (int i = 0; i < 26; i++) nxt[1][i] = 0;
 67     }
 68 
 69     void extend(int c) {
 70         int u = ++tot, v = last;
 71         len[u] = len[v] + 1;
 72         num[u] = 1;
 73         for (; v && !nxt[v][c]; v = fail[v]) nxt[v][c] = u;
 74         if (!v) fail[u] = 1, sz[1]++;
 75         else if (len[nxt[v][c]] == len[v] + 1) fail[u] = nxt[v][c], sz[nxt[v][c]]++;
 76         else {
 77             int now = ++tot, cur = nxt[v][c];
 78             len[now] = len[v] + 1;
 79             memcpy(nxt[now], nxt[cur], sizeof(nxt[cur]));
 80             fail[now] = fail[cur];
 81             fail[cur] = fail[u] = now;
 82             for (; v && nxt[v][c] == cur; v = fail[v]) nxt[v][c] = now;
 83             sz[now] += 2;
 84         }
 85         last = u;
 86         //return len[last] - len[fail[last]];//多添加一个子串所产生不同子串的个数
 87     }
 88 
 89     void get_num() {// 每个节点子串出现的次数
 90         for (int i = 1; i <= tot; i++) X[len[i]]++;
 91         for (int i = 1; i <= tot; i++) X[i] += X[i - 1];
 92         for (int i = 1; i <= tot; i++) Y[X[len[i]]--] = i;
 93         for (int i = tot; i >= 1; i--) num[fail[Y[i]]] += num[Y[i]];
 94     }
 95 
 96     void get_maxx(int n) {// 长度为x的子串出现次数最多的子串的数目
 97         get_num();
 98         for (int i = 1; i <= tot; i++) maxx[len[i]] = max(maxx[len[i]], num[i]);
 99     }
100 
101     void get_sum() {// 该节点后面所形成的自字符串的总数
102         get_num();
103         for (int i = tot; i >= 1; i--) {
104             sum[Y[i]] = 1;
105             for (int j = 0; j <= 25; j++)
106                 sum[Y[i]] += sum[nxt[Y[i]][j]];
107         }
108     }
109 
110     void get_subnum() {//本质不同的子串的个数
111         subnum = 0;
112         for (int i = 1; i <= tot; i++) subnum += len[i] - len[fail[i]];
113     }
114 
115     void get_sublen() {//本质不同的子串的总长度
116         sublen = 0;
117         for (int i = 1; i <= tot; i++) sublen += 1LL * (len[i] + len[fail[i]] + 1) * (len[i] - len[fail[i]]) / 2;
118     }
119 
120     void get_sa() { //获取sa数组
121         for (int i = 1; i <= tot; i++) c[len[i]]++;
122         for (int i = 1; i <= tot; i++) c[i] += c[i - 1];
123         for (int i = tot; i >= 1; i--) sa[c[len[i]]--] = i;
124     }
125 
126 
127     int minn[maxn << 1], mx[maxn << 1];//多个串的最长公共子串
128     //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
129 
130     void match(char s[]) {
131         mem(mx, 0);
132         int n = strlen(s), p = 1, maxlen = 0;
133         for (int i = 0; i < n; i++) {
134             int c = s[i] - 'a';
135             if (nxt[p][c]) p = nxt[p][c], maxlen++;
136             else {
137                 for (; p && !nxt[p][c]; p = fail[p]);
138                 if (!p) p = 1, maxlen = 0;
139                 else maxlen = len[p] + 1, p = nxt[p][c];
140             }
141             mx[p] = max(mx[p], maxlen);
142         }
143         for (int i = tot; i; i--)
144             mx[fail[i]] = max(mx[fail[i]], min(len[fail[i]], mx[i]));
145         for (int i = tot; i; i--)
146             if (minn[i] == -1 || minn[i] > maxx[i]) minn[i] = mx[i];
147     }
148 
149     void get_kth(int k) {//求出字典序第K的子串
150         int pos = 1, cnt;
151         string s = "";
152         while (k) {
153             for (int i = 0; i <= 25; i++) {
154                 if (nxt[pos][i] && k) {
155                     cnt = nxt[pos][i];
156                     if (sum[cnt] < k) k -= sum[cnt];
157                     else {
158                         k--;
159                         pos = cnt;
160                         s += (char) (i + 'a');
161                         break;
162                     }
163                 }
164             }
165         }
166         cout << s << endl;
167     }
168 } sam;
169 
170 char s[maxn];
171 
172 int main() {
173 #ifndef ONLINE_JUDGE
174     FIN;
175 #endif
176     sam.init();
177     sfs(s + 1);
178     int n = strlen(s + 1);
179     for (int i = 1; i <= n; i++) sam.extend((s[i] - 'a'));
180     sam.get_maxx(n);
181    // fuck(n);
182     for (int i = 1; i <= n; i++) printf("%d\n", sam.maxx[i]);
183 #ifndef ONLINE_JUDGE
184     cout << "Totle Time : " << (double) clock() / CLOCKS_PER_SEC << "s" << endl;
185 #endif
186     return 0;
187 }
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