1040 count the number of words

1040 Count the number of words

2001 NOIP National League Improvement Group

time limit: 1 sspace limit: 128000 KBtitle level: Golden Gold SolvingTitle Description Description

Give a letter string with a length of not more than 200, consisting of a lowercase English letter (contract; The string is entered in 20 letters per line and guaranteed to be 20 each). This letter is required to be divided into K-parts (1<k<=40), and the number of words contained in each copy is the largest (the words contained in each copy can be partially overlapped). After a word is selected, its first letter cannot be used again. For example, the string this can contain this and is, after selecting this cannot contain th) (Administrator Note: Here can not be used to refer to the position, not the letter itself. For example, thisis can be counted as including 2 is).
The word is given in a dictionary of no more than 6 words.
The maximum number of outputs required.

Enter a description Input Description

First behavior a positive integer (0<n<=5) indicates that there are n sets of test data
The first row of each group has two positive integers (p,k)
P indicates the number of lines of string;
K is divided into k parts.
The next P-line, each line has 20 characters.
Then there is a positive integer s, which indicates the number of words in the dictionary. (1<=s<=6)
The next S-line, each line has one word.

Output description Output Description

Each row is an integer that corresponds to the corresponding result for each set of test data.

Sample input Sample Input

1
1 3
Thisisabookyouareaoh
4
Is
A
Ok
Sab

Sample output Sample Output

7

Data range and Tips Data Size & Hint

This/isabookyoua/reaoh

Category labels Tags Click here to expandPartition DP Dynamic Planning Continental region Noip National League increase Group 2001

AC Code:

/*Solution : Divide the string ss[0..len-1] into the optimal value of the K section, consider dividing the first I characters into the optimal value F (i,j) =max{f (i-x,j-1) + the number of words in X characters} (I>=j and x>=) of J parts. 1,I-X>=J-1) is 1<=x<i-j for the interval [ii.. JJ] contains the number of words, one by one to S[kk] (II<=KK<=JJ) The beginning of the word can be,*/#include<cstdio>#include<iostream>#include<cstring>using namespacestd;#defineN 211intn,p,k,s,f[n][ A],len;Charss[n],tt[n],w[ A][n]; voidFirst () {memset (F,0,sizeof(f));//develop a good habit of clearing 0Memset (W,0,sizeof(w)); memset (TT,0,sizeof(TT)); memset (SS,0,sizeof(ss)); scanf ("%d%d",&p,&k); Len=0;  for(intI=0; i<p;i++) {scanf ("%s", TT);  for(intj=0; tt[j];j++) {Ss[len]=tt[j];len++;//string Pressure stack}} Ss[len]=' /'; scanf ("%d", &s);//number of words     for(intI=0; i<s;i++) scanf ("%s", W[i]);//Word}intdpintXinty) {    intans=0, cur;  for(inti=x;i<=y;i++){         for(intj=0; j<s;j++){             for(cur=0; w[j][cur];cur++)if(i+cur>y| | W[j][cur]!=ss[i+cur]) Break; if(w[j][cur]==' /') {ans++; Break;//sentence The last of these words            }            }    }        returnans;}intMain () {scanf ("%d",&N);  while(n--) {first ();  for(intI=1; i<=len;i++) f[i][1]=DP (0, I-1);//Initialize F         for(intj=2; j<=k;j++) {//Note I j Order             for(inti=j;i<=len;i++){                 for(intx=1; x<i-j;x++) {//Enumeration Sub-method                    intTMP=DP (i-x,i-1); F[I][J]=max (f[i][j],f[i-x][j-1]+tmp); }}} printf ("%d\n", F[len][k]); }    return 0;}

1040 Count Words