poj3974 Palindrome

题意：求一个字符串的最长回文子串

题解：是一个Manacher模板题，为了统一奇偶，先预处理在字符间添加间隔，使字符串长度变为偶数，例如"abc"添加分隔符后变成"$#a#b#c#"，剩下的就很简单了，这个题也可以用哈希做。就时间复杂度来看Manacher明显比哈希快。

Manacher代码

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<string.h>
 4 #include<string>
 5 using namespace std;
 6 
 7 int p[2000020];//p数组需要开至少字符串两倍的空间
 8 char s[1010100];
 9 
10 int Manacher(char str[])
11 {
12     memset(p,0,sizeof(p));
13     string s="$#";
14     int len=strlen(str);
15     for(int i=0;i<len;i++){
16         s+=str[i];
17         s+='#';
18     }
19     int mx=0,id=0,reslen=0,rescenter=0;
20     len=s.length();
21     for(int i=1;i<=len;i++)
22     {
23         p[i]=mx>i?min(p[2*id-i],mx-i):1;//
24         while(s[i+p[i]]==s[i-p[i]])p[i]++;
25         if(mx<i+p[i])
26         {
27             mx=i+p[i];
28             id=i;
29         }
30         if(reslen<p[i])
31         {
32             reslen=p[i];
33             rescenter=i;
34         }
35     }
36     return reslen-1;
37 }
38 
39 
40 int main()
41 {
42     int t=1;
43     while(~scanf("%s",s)){
44         if(strcmp(s,"END")==0) break;
45         printf("Case %d: %d\n",t++,Manacher(s));
46     }
47     return 0;
48 }

哈希代码

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<string.h>
 4 #include<math.h>
 5 #include<iostream>
 6 using namespace std;
 7 
 8 const int maxn = 2e6+20;
 9 const int base = 131;　　　　　//不易冲突 10 typedef unsigned long long ULL;
11 ULL f[maxn], f2[maxn], t[maxn];
12 
13 char s[maxn];
14 ULL solve(ULL h[],int l, int r)
15 {
16     return h[r] - h[l-1] * t[r-l+1];
17 }
18 
19 int main()
20 {
21     int T = 1;
22     while(scanf("%s", s+1) != EOF) {
23         int len = strlen(s+1);
24         if(!strcmp(s+1, "END"))
25             break;
26         for(int i = 2*len; i > 0; i-=2) {
27             s[i] = s[i/2];
28             s[i-1] = 'z'+1;
29         }
30         len *= 2;
31         f[0] = f2[0] = 0;
32         t[0] = 1;
33         for(int i = 1, j = len; i <= len; i++, j--) {
34             f[i] = f[i-1]*base + (s[i] - 'a' + 1);
35             f2[i] = f2[i-1]*base + (s[j] - 'a' + 1);
36             t[i] = t[i-1] * base;
37         }
38         int l, r;
39         int ans = 0;
40         int mid;
41         for(int i = 1; i <= len; i++) {
42             l = 0, r = min(i-1, len-i);
43             while(l < r) {
44                 mid = (l+r+1)>>1;
45                 if(solve(f,i-mid,i-1) != solve(f2,len-(i+mid)+1,len-(i+1)+1))
46                     r = mid-1;
47                 else l = mid;
48             }
49             if(s[i-l] <= 'z')
50                 ans = max(ans, l+1);
51             else
52                 ans = max(ans, l);
53         }
54         printf("Case %d: %d\n", T++, ans);
55     }
56     return 0;
57 }