CSU 1328 近似回文词（2013湖南省程序设计竞赛A题）

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 using namespace std;
 6 const int maxn = 1005;
 7 char tt[maxn],str[maxn];
 8 int loc[maxn],K;
 9 
10 char chto(char c)
11 {
12     return c >= ‘A‘ && c <= ‘Z‘? (c+‘a‘-‘A‘):c;
13 }
14 void StrCmp(char* str,int i,int j,int& left,int& right)
15 {
16     int k = 0,len = strlen(str);
17     while(i >= 0 && j < len)
18     {
19         if(str[i] != str[j]) k++;
20         if(k > K)
21         break;
22         i--,j++;
23     }
24     left = i+1;
25     right = j - 1;
26 }
27 int kuoleft(int l)
28 {
29     l--;
30     while(l >= 0 && !((tt[l] >= ‘A‘ && tt[l] <= ‘Z‘) || (tt[l] >= ‘a‘ && tt[l] <= ‘z‘)))
31     l--;
32     return l+1;
33 }
34 int kuoright(int r)
35 {
36     r++;
37     int len = strlen(tt);
38     while(r < len && !((tt[r] >= ‘A‘ && tt[r] <= ‘Z‘) || (tt[r] >= ‘a‘ && tt[r] <= ‘z‘)))
39     r++;
40     return r - 1;
41 }
42 int main()
43 {
44     int kase = 1;
45     while(scanf("%d",&K)!=EOF)
46     {
47         getchar();
48         gets(tt);
49         int f = strlen(tt),len = 0;
50         for(int i = 0;i < f;++i)
51         if((tt[i] >= ‘A‘ && tt[i] <= ‘Z‘) || (tt[i] >= ‘a‘ && tt[i] <= ‘z‘))
52         {
53             str[len] = chto(tt[i]);
54             loc[len] = i;
55             len++;    //记录这个字符在原来的子串中对应的位置 
56         }
57         str[len] = NULL;
58         int s = 0,L = 0;
59         for(int l = 0;l < len;++l)
60         {
61             int left = 0,right = 0;
62             StrCmp(str,l-1,l+1,left,right);
63             int ll = loc[left],rr = loc[right];
64             int lt = rr - ll + 1;
65             if(lt > L) L = lt,s = ll;
66             else if(lt == L && ll < s) s = ll;
67             if(l < len -1)
68             {
69                 left = right = 0;
70                 StrCmp(str,l,l+1,left,right);
71                 ll = loc[left],rr = loc[right];
72                 lt = rr - ll + 1;
73                 if(lt > L) L = lt,s = ll;
74                 else if(lt == L && ll < s) s = ll;
75             }
76         }
77         printf("Case %d: %d %d\n",kase++,L,s+1);
78     }
79     return 0;
80 }