dp topic title record

1. The number of division 2 stars

https://ac.nowcoder.com/acm/problem/16695

. 1 #include <bits / STDC ++ H.>
 2  the using  namespace STD;
 . 3  #define LL Long Long
 . 4  #define EPS-1E. 6
 . 5  int ANS = 0 ;
 . 6  int n-, K;
 . 7  void DFS ( int ID, int SUM, int Val)
 . 8  {
 . 9      / * IF (K-ID. 1 && == + SUM Val <= n-) {// the presence of the last digit n-sum> = val set which is incremented number 
 10          ANS ++;
 . 11          return;
 12 is      }
 13 is      for (int I = Val; I <= n-; I ++) {
14         if(sum + i >n)
15             break;
16         dfs( id + 1, sum + i, i);
17     }*/ 
18    if(id==k&&sum==n)
19     {
20         ans++;
21         return;
22      }
23       
24     if(id>=k||sum>n)
25     return;
26     
27     for(int i = val; i <= n; i++){
28         if(sum+i>n)
29         break;
30         dfs(id+1,sum+i,i);
31     }
32 }
33  
34 int main()
35 {
36     while(scanf("%d %d", &n, &k) != EOF){
37         ans = 0;
38         dfs(0, 0, 1);
39         printf("%d\n", ans);
40     }
41     return 0;
42 }

View Code

 

2. The word Solitaire 3 stars

. 1 #include <bits / STDC ++ H.>
 2  the using  namespace STD;
 . 3  
. 4  struct Node {
 . 5      char S [ 22 is ];
 . 6      int len;
 . 7      int V; // number of times selected 
. 8 } C [ 22 is ];
 . 9  int n, MAXN;
 10  
. 11  void DFS ( int X, int len) { // current word length Long current number 
12 is      for ( int I = . 1 ; I <= n; I ++) {    // can be selected from the group of n word
13 is          IF (C [I] .v < 2 ) // number of times smaller than the selected 2 
14          {
 15              for ( int J = 0 ; J <C [X] .LEN; J ++) {    // a word is determined before and currently selected word have equal areas 
16                  IF (C [X] .s [J] == C [I] .s [ 0 ]) {   
 . 17                  int K = . 1 , T = . 1 , L;
 18 is                  for (L J + = . 1 ; L <C [X] && .LEN K <C [I] .LEN; K ++, ++ L) {
 . 19                      IF ! (C [X] .s [L] = C [I] .s [ K])
 20 is                      {
 21 is                          T = 0;    BREAK ;
 22 is                      }
 23 is                  }
 24                  IF (L = C [x] .LEN!)   // DESCRIPTION word x i is a word comprising 
25                  T = 0 ;
 26 is                  IF (T)   // Description word x i is connected to the word and not contained 
27                  {
 28                      C [I] .v ++ ;
 29                      DFS (I, C + len [I] .len- K);  
 30                      C [I] .v-- ;
 31 is                  }                
 32                  }
 33 is              }
 34 is         }
 35      }
 36      MAXN = max (MAXN, len);
 37 [  }
 38 is  
39  int main () {
 40      CIN n->>; // number of words 
41 is      for ( int I = . 1 ; I <= n-; I ++ ) {
 42 is      CIN C >> [I] .s; // first word content 
43 is      C [I] .LEN = strlen (C [I] .s); // first word length 
44 is      }
 45      CIN >> C [ 0 ] .s; // idiom initials     
46 is      C [ 0 ] = .LEN strlen (C [ 0 ] .s);
47     
48     dfs(0,c[0].len);
49     cout<<maxn<<endl;
50 }

View Code

Using dfs of: searching for a maximum value of the program 
program composed by several x, x there are n kinds of programs, for x in dfs 
void dfs ( int x) 
{ 
for ( int I = 0 ; I <n; I ++ ) 
.. . 
}