UVA 11584 "Partitioning by Palindromes"（DP+Manacher）

Portal

 

• the meaning of problems　　

• ideas a

　　Defined dp [i] represents the minimum number of 0 ~ i of the division;

　　First of all, to solve the array of horse-drawn vehicles palindrome radius algorithm;

　　For the i-th character S _i , traversing j (0 ≤ j <i) , j is determined in the center of the maximum palindrome palindromic sequence contains S _i ;

　　If included, dp [i] = min {dp [i], dp [2 * ji-1] +1};

•Code

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e3+50;
 4 
 5 char t[maxn];
 6 int r[maxn<<1];
 7 
 8 struct Manacher
 9 {
10     char s[maxn<<1];
11     void Init(char *ss,int len)
12     {
13         int index=0;
14         s[index++]='#';
15         for(int i=0;i < len;++i)
16         {
17             s[index++]=ss[i];
18             s[index++]='#';
19         }
20         s[index]='\0';
21     }
22     void mana(char *ss)
23     {
24         Init(ss,strlen(ss));
25         int len=strlen(s);
26         int R=-1;
27         int C;
28         for(int i=0;i < len;++i)
29         {
30             r[i]=R > i ? min(R-i+1,r[2*C-i]):1;
31             for(;i-r[i] >= 0 && i+r[i] < len && s[i-r[i]] == s[i+r[i]];r[i]++);
32             if(i+r[i] > R)
33             {
34                 R=i+r[i]-1;
35                 C=i;
36             }
37         }
38     }
39 }_mana;
40 
41 int dp[maxn];
42 int Solve()
43 {
44     _mana.mana(t);
45 
46     dp[0]=1;
47     int len=strlen(t);
48     for(int i=1;i < len;++i)
49     {
50         dp[i]=dp[i-1]+1;
51         for(int J = 0 ; J <= 2 * i; ++ J)
 52 is          {
 53 is              
54 is              /// position t of the i-th character in the preprocessed array s is i +. 1 * 2
 55              /// because there may be a case where even palindromic, it is necessary to use the '#'
 56              /// directly judge the array corresponding to the maximum of j in the palindrome j s + r [j] is included. 1 + I * 2
 57 is              /// If so, then find the 2 * i + 1 to j is the center point of symmetry of 2 * j- (2 * i +
 1) 58              /// Analyzing 2 * j- (2 * i + 1) in a position corresponding to t a front position (2 * j- (2 * i + 1)) / 2-1 is in [0, len-1] in the range of
 59              /// If the updated DP [I] 
60              int CUR = J + R & lt [J];
 61 is              int index = ( 2 * J- 2 * I- . 1 ) / 2 -1;
62             if(2*i+1 < cur)
63                 dp[i]=min(dp[i],1+(index >= 0 ? dp[index]:0));
64         }
65     }
66     return dp[len-1];
67 }
68 int main()
69 {
70 //    freopen("C:\\Users\\hyacinthLJP\\Desktop\\in&&out\\contest","r",stdin);
71     int test;
72     scanf("%d",&test);
73     while(test--)
74     {
75         scanf("%s",t);
76         printf("%d\n",Solve());
77     }
78     return 0;
79 }

View Code 

• thinking two (reference from zishu)

　　Defined dp [i] denotes 0 ~ i into a minimum number of palindromic sequence, the dp [i] = min {d [j] +1 | j ≤ i && t [j + 1, ...., i]} is a palindromic sequence;

•code

. 1 #include <bits / STDC ++ H.>
 2  the using  namespace STD;
 . 3  const  int MAXN = 1E3 + 50 ;
 . 4  
. 5  char T [MAXN];
 . 6  char S [MAXN << . 1 ];
 . 7  BOOL IsPAL [MAXN] [MAXN] ;
 . 8  int DP [MAXN];
 . 9  
10  void the Init () /// O (^ n-2) pretreatment of the t [i, .., j] if the palindromic sequence 
. 11  {
 12 is      int len = strlen (T) ;
 13 is      for ( int I = 0 ; I <len; ++ I)
14         for(int j=0;j < len;++j)
15             isPal[i][j]=false;
16     int index=0;
17     s[index++]='#';
18     for(int i=0;i < len;++i)
19     {
20         s[index++]=t[i];
21         s[index++]='#';
22     }
23     s[index]='\0';
24 
25     for(int i=0;i < index;++i)
26     {
27         int r=0;
28         while(i-r >= 0 && i+r < index && s[i-r] == s[i+r])
29         {
30             if((i-r)&1)
31                 isPal[(i-r)/2][(i+r)/2]=true;
32             r++;
33         }
34     }
35 }
36 
37 int Solve()
38 {
39     Init();
40     int len=strlen(t);
41     dp[0]=1;
42     for(int i=1;i < len;i++)
43     {
44         dp[i]=dp[i-1]+1;
45         for(int j=0;j < i;++j)
46             if(isPal[j][i])
47                 dp[i]=min(dp[i],1+(j > 0 ? dp[j-1]:0));
48     }
49     return dp[len-1];
50 }
51 int main()
52 {
53 //    freopen("C:\\Users\\hyacinthLJP\\Desktop\\in&&out\\contest","r",stdin);
54     int test;
55     scanf("%d",&test);
56     while(test--)
57     {
58         scanf("%s",t);
59         printf("%d\n",Solve());
60     }
61     return 0;
62 }

View Code