Uestc_judgment Day Cdoj 11

Today is the judgment Day. The world was ending and all man would pay for their guilt and sin. Now the Almighty God has a long string, and some people ' s name. He'll take some letters from the string and use them to spell out those people ' s name. What's the maximum number of the names that can being spelled, with each name being spelled at most once?

Input

The first line of input contains a numberT, indicating the number of test cases. (T≤ ) For each test case, the first line contains a strings, which is the string the God has. The following line contains a numberNThe number of people ' s names. ThenNLines follow, each with a string indicating the names. There'll be is no more thanTenNames in all test case, the length of stringswould not exceedAnd the length of each name would not exceed . All the strings contains lowercase letters a z from to only.

Output

For each case, output first Case #i: . (i is the number of the "test case" from 1 to T). Then output a single number, as the maximum number of the people whose name can be spelled.

Sample Input and output

Sample Input	Sample Output
2abc3aabbccaabbc3abcaabb	Case #1:0Case #2:2

Sourcethe 11th UESTC Programming Contest FinalProblem Solving Report:first sweep out each name to how many characters, after the violent enumeration spell the way and update ans

#include <iostream>#include<cstring>#include<algorithm>using namespacestd;intletter[ -+5];intcost[Ten+3][ -+5];intans;Charbuffer[100000+ -];intMainintARGC,Char*argv[]) { intcase,t=1; scanf ("%d%*c",&Case );  while(case--) {memset (letter,0,sizeof(letter)); memset (Cost,0,sizeof(cost)); intLen,n; Gets (buffer); Len=strlen (buffer);  for(inti =0; i < Len; ++i) letter[buffer[i]-'a']++; scanf ("%d%*c",&N);  for(inti =0; I < n; ++i) {gets (buffer); Len=strlen (buffer);  for(intj =0; J < Len; ++j) Cost[i][buffer[j]-'a']++; }    intallcost[ -+5]; Ans=0;  for(inti =1; I < (1&LT;&LT;N); ++i) {memset (Allcost,0,sizeof(Allcost)); intNewans =0;  for(intj =0; J < N; ++j)if((i >> J) &1) {Newans++;  for(intK =0; K < -; ++k) Allcost[k]+=Cost[j][k]; }        intFlag =1;  for(intj =0; J < -; ++j)if(Allcost[j] >Letter[j]) {Flag=0;  Break; }        if(flag) ans=Max (Ans,newans); } printf ("Case #%d:%d\n", t++, ans); } return 0;}

Uestc_judgment Day Cdoj 11