Partitioning by Palindromes UVA - 11584 简单dp

题目:题目链接

思路:预处理出l到r为回文串的子串,然后如果j到i为回文串,dp[i] = min(dp[i], dp[j] + 1)

AC代码:

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 #include 
 6 #include 
 7 #include <string>
 8 #include <set>
 9 #include 
10 #include 
11 #include 
12 #include 
13 #include 
14 
15 #define FRER() freopen("in.txt", "r", stdin)
16 #define FREW() freopen("out.txt", "w", stdout)
17 #define INF 0x3f3f3f3f
18 
19 using namespace std;
20 
21 const int maxn = 1000 + 5;
22 
23 int n, len, l, r;
24 char str[maxn];
25 
26 bool check[maxn][maxn];
27 int dp[maxn];
28 
29 int main()
30 {
31     //FRER();
32     scanf("%d", &n);
33     while(n--) {
34         scanf("%s", str);
35         len = strlen(str);
36         memset(check, 0, sizeof(check));
37 
38         for(int i = 0; i < len; ++i) {
39             l = r = i;
40             while(l >= 0 && r < len) {
41                 if(str[l] == str[r]) {
42                     check[l + 1][r + 1] = 1;
43                     --l; ++r;
44                 }
45                 else break;
46             }
47             l = i;
48             r = i + 1;
49             
50             while(l >= 0 && r < len) {
51                 if(str[l] == str[r]) {
52                     check[l + 1][r + 1] = 1;
53                     --l; ++r;
54                 }
55                 else break;
56             }
57         }
58 
59         memset(dp, 0, sizeof(dp));
60         for(int i = 1; i <= len; ++i) {
61             dp[i] = dp[i - 1] + 1;
62             for(int j = 1; j < i; ++j) {
63                 if(check[j][i])
64                     dp[i] = min(dp[i], dp[j - 1] + 1);
65             }
66         }
67 
68         printf("%d\n", dp[len]);
69     }
70     return 0;
71 }

 

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