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  • HDU 4416 Good Article Good sentence

    Good Article Good sentence

    Time Limit: 3000ms
    Memory Limit: 32768KB
    This problem will be judged on HDU. Original ID: 4416
    64-bit integer IO format: %I64d      Java class name: Main
     
    In middle school, teachers used to encourage us to pick up pretty sentences so that we could apply those sentences in our own articles. One of my classmates ZengXiao Xian, wanted to get sentences which are different from that of others, because he thought the distinct pretty sentences might benefit him a lot to get a high score in his article.
    Assume that all of the sentences came from some articles. ZengXiao Xian intended to pick from Article A. The number of his classmates is n. The i-th classmate picked from Article Bi. Now ZengXiao Xian wants to know how many different sentences she could pick from Article A which don't belong to either of her classmates?Article. To simplify the problem, ZengXiao Xian wants to know how many different strings, which is the substring of string A, but is not substring of either of string Bi. Of course, you will help him, won't you?
     

    Input

    The first line contains an integer T, the number of test data. 
    For each test data
    The first line contains an integer meaning the number of classmates.
    The second line is the string A;The next n lines,the ith line input string Bi.
    The length of the string A does not exceed 100,000 characters , The sum of total length of all strings Bi does not exceed 100,000, and assume all string consist only lowercase characters 'a' to 'z'.
     

    Output

    For each case, print the case number and the number of substrings that ZengXiao Xian can find.
     

    Sample Input

    3
    2
    abab
    ab
    ba
    1
    aaa
    bbb
    2
    aaaa
    aa
    aaa

    Sample Output

    Case 1: 3
    Case 2: 3
    Case 3: 1

    Source

     
    题意:给一个字符串S和一系列字符串T1~Tn,问在S中有多少个不同子串满足它不是T1~Tn中任意一个字符串的子串
    解题:后缀自动机
     
     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 typedef long long LL;
     4 const int maxn = 100010;
     5 int sa[maxn<<1],c[maxn];
     6 struct node{
     7     int son[26],f,len,deep;
     8     void init(){
     9         memset(son,-1,sizeof son);
    10         deep = len = 0;
    11         f = -1;
    12     }
    13 };
    14 struct SAM{
    15     node e[maxn<<1];
    16     int tot,last;
    17     int newnode(int len = 0){
    18         e[tot].init();
    19         e[tot].len = len;
    20         return tot++;
    21     }
    22     void init(){
    23         tot = last = 0;
    24         newnode();
    25     }
    26     void add(int c){
    27         int p = last,np = newnode(e[p].len + 1);
    28         while(p != -1 && e[p].son[c] == -1){
    29             e[p].son[c] = np;
    30             p = e[p].f;
    31         }
    32         if(p == -1) e[np].f = 0;
    33         else{
    34             int q = e[p].son[c];
    35             if(e[p].len + 1 == e[q].len) e[np].f = q;
    36             else{
    37                 int nq = newnode();
    38                 e[nq] = e[q];
    39                 e[nq].len = e[p].len + 1;
    40                 e[q].f = e[np].f = nq;
    41                 while(p != -1 && e[p].son[c] == q){
    42                     e[p].son[c] = nq;
    43                     p = e[p].f;
    44                 }
    45             }
    46         }
    47         last = np;
    48     }
    49     void update(char *str){
    50         int ln = 0,p = 0;
    51         for(int i = 0; str[i]; ++i){
    52             int c = str[i] - 'a';
    53             if(e[p].son[c] != -1){
    54                 p = e[p].son[c];
    55                 e[p].deep = max(e[p].deep,++ln);
    56             }else{
    57                 while(p != -1 && e[p].son[c] == -1) p = e[p].f;
    58                 if(p == -1) p = ln = 0;
    59                 else{
    60                     ln = e[p].len + 1;
    61                     p = e[p].son[c];
    62                     e[p].deep = max(e[p].deep,ln);
    63                 }
    64             }
    65         }
    66     }
    67     LL count(){
    68         LL ret = 0;
    69         for(int i = tot-1; i > 0; --i){
    70             int v = sa[i];
    71             if(e[v].deep > 0){
    72                 e[e[v].f].deep = max(e[e[v].f].deep,e[v].deep);
    73                 if(e[v].deep < e[v].len) ret += e[v].len - e[v].deep;
    74             }else ret += e[v].len - e[e[v].f].len;
    75         }
    76         return ret;
    77     }
    78 }sam;
    79 char str[maxn];
    80 int main(){
    81     int kase,m,cs = 1;
    82     scanf("%d",&kase);
    83     while(kase--){
    84         scanf("%d%s",&m,str);
    85         sam.init();
    86         for(int i = 0; str[i]; ++i) sam.add(str[i] - 'a');
    87         memset(c,0,sizeof c);
    88         for(int i = 0; i < sam.tot; ++i) c[sam.e[i].len]++;
    89         for(int i = 1,len = strlen(str); i <= len; ++i) c[i] += c[i-1];
    90         for(int i = sam.tot-1; i >= 0; --i) sa[--c[sam.e[i].len]] = i;
    91         while(m--){
    92             scanf("%s",str);
    93             sam.update(str);
    94         }
    95         printf("Case %d: %I64d
    ",cs++,sam.count());
    96     }
    97     return 0;
    98 }
    View Code
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  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4805447.html
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