poj 1159 Palindrome - 动态规划

 1 /**
 2  * poj
 3  * Problem#1159
 4  * Accepted
 5  * Time:782ms
 6  * Memory:748k
 7  */
 8 #include<iostream>
 9 #include<sstream>
10 #include<cstdio>
11 #include<cmath>
12 #include<cstdlib>
13 #include<cstring>
14 #include<cctype>
15 #include<queue>
16 #include<set>
17 #include<map>
18 #include<stack>
19 #include<vector>
20 #include<algorithm>
21 #ifdef    WIN32
22 #define    AUTO "%I64d"
23 #else
24 #define AUTO "%lld"
25 #endif
26 using namespace std;
27 typedef bool boolean;
28 #define smin(a, b) (a) = min((a), (b))
29 #define smax(a, b) (a) = max((a), (b))
30 template<typename T>
31 inline void readInteger(T& u){
32     char x;
33     int aFlag = 1;
34     while(!isdigit((x = getchar())) && x != ‘-‘);
35     if(x == ‘-‘){
36         aFlag = -1;
37         x = getchar();
38     }
39     for(u = x - ‘0‘; isdigit((x = getchar())); u = u * 10 + x - ‘0‘);
40     ungetc(x, stdin);
41     u *= aFlag;
42 }
43 
44 template<typename T>class Matrix{
45     public:
46         T *p;
47         int lines;
48         int rows;
49         Matrix():p(NULL){    }
50         Matrix(int rows, int lines):lines(lines), rows(rows){
51             p = new T[(lines * rows)];
52         }
53         T* operator [](int pos){
54             return (p + pos * lines);
55         }
56 };
57 #define matset(m, i, s) memset((m).p, (i), (s) * (m).lines * (m).rows)
58 
59 int n;
60 char* str;
61 
62 inline void init() {
63     readInteger(n);
64     str = new char[(const int)(n + 3)];
65     getchar();
66     gets(str + 1);
67 }
68 #define idx(i)    ((i < 0) ? (i + 3) : (i))
69 
70 //Matrix<int> f;
71 int f[3][5005];
72 inline void solve() {
73 //    f = Matrix<int>(3, n + 1);
74     for(int i = 1; i <= n; i++)
75         f[0][i] = 0;
76     for(int i = 1; i < n; i++)
77         f[1][i] = (str[i] == str[i + 1]) ? (0) : (1);
78     int t = 2;
79     for(int k = 2; k < n; k++) {
80         for(int i = 1; i + k <= n; i++) {
81             int j = i + k;
82             if(str[i] == str[j])    f[t][i] = f[idx(t - 2)][i + 1];
83             else    f[t][i] = min(f[idx(t - 1)][i], f[idx(t - 1)][i + 1]) + 1;
84         }
85         t = (t == 2) ? (0) : (t + 1);
86     }
87     printf("%d", f[idx(t - 1)][1]);
88 }
89 
90 int main() {
91     init();
92     solve();
93     return 0;
94 }