UVa 10887:concatenation of Languages

Link:

Uva:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&itemid=8&category=24&page=show_ problem&problem=1828

Type: Hash table

Original title:

A language is a set of strings. And the concatenation of two languages is the set of all strings this are formed by concatenating the strings of the Secon D language at the "strings of the" the "the" the "the" the "the".

For example, if we have two language A and B such that:

A = {cat, dog, mouse}

B = {rat, bat}

The concatenation of A and B would be:

C = {Catrat, catbat, Dograt, Dogbat, Mouserat, Mousebat}

Given Two languages your task is ' only ' to count the # of strings in the concatenation of the two.

Sample input;

2

3 2

Cat

Dog

Mouse

Rat

Bat

1 1

Abc

Cab

Sample output:

Case 1:6

Case 2:1

The main effect of the topic:

There are word sets a and set B, and then there are these two sets that make up a new compound, where a is the first half of the compound, and the second part of B. Give the A,b two collection, ask how many match words can be combined?

Analysis and Summary:

This feeling is UVA 10391-compound Words's brother, quite like.

Or use hash to judge heavy. Try to join a hash table without a new word, and add 1 if you succeed.

This problem is because of the misuse of memcmp and memspy to handle strings and WA several times.

There is a very pit dad, read the word to use gets, because there may be empty lines of words ....

* * UVA 10887-concatenation of Languages * time:0.404s (UVA) * author:d_double/#include <iostr  
     
eam> #include <cstring> #include <cstdio> #define MAXN 1503 using namespace std;  
Char a[maxn][12], c[maxn*maxn][24];  
const int hashsize = MAXN*MAXN;  
     
int head[hashsize], next[hashsize], rear, ans;  inline void init () {rear=1; ans=0; memset (head, 0, sizeof);  
    } inline int hash (char *str) {int seed=131, v=0;  
    while (*str) v = v*seed + (*str++);  
Return (V & 0x7fffffff)%hashsize;  
    int Try_to_insert (int s) {int h = hash (c[s]);  
    int u = head[h];  
        while (U) {if (strcmp (C[u], c[s]) ==0) return 0;  
    U = next[u];  
    } Next[s] = Head[h];  
    HEAD[H] = s;  
return 1;  
    int main () {int T, M, N, Cas=1;  
    scanf ("%d%*c", &t);  
        while (t--) {scanf ("%d%d%*c", &m,&n); for (int i=0; i<m; ++i) gets (a[i));  
     
        Init ();  
        Char str[12], link_word[24];  
            for (int j=0; j<n; ++j) {gets (str);  
                for (int i=0; i<m; ++i) {strcpy (Link_word, a[i]);  
                strcat (Link_word, str);  
                strcpy (C[rear], Link_word);  
            if (Try_to_insert (rear)) {++rear; ++ans;}  
    } printf ("Case%d:%d\n", cas++, ans);  
return 0; }