UVA11584 划分成回文串

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=105116#problem/B

紫书275

题意:输入一个字符，最少能划分几个回文串

分析：预处理一下，判断i，j是否为回文串；动态分析求解，dp[i] = dp[i - 1] + 1,假设i单独成为一个回文串，然后在往前找，如果j到i是回文，dp[i] = min(dp[i], dp[j - 1] + 1)

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 using namespace std;
 6 const int MAX = 1000 + 10;
 7 const int INF = 0x3f3f3f3f;
 8 char s[MAX];
 9 int dp[MAX],len,is_h[MAX][MAX];
10 void judge_huiwen()
11 {
12     int x,y;
13     memset(is_h, 0, sizeof(is_h));
14     for(int i = 1; i <= len; i++)
15     {
16         x = i - 1;
17         y = i + 1;
18         while(x > 0 && y <= len && s[x] == s[y])
19         {
20             is_h[x][y] = 1;
21             x--;
22             y++;
23         }
24         x = i;
25         y = i + 1;
26         while(x > 0 && y <= len && s[x] == s[y])
27         {
28             is_h[x][y] = 1;
29             x--;
30             y++;
31         }
32     }
33 }
34 int main()
35 {
36     int test;
37     scanf("%d", &test);
38     while(test--)
39     {
40         scanf("%s", s + 1);
41         len = strlen(s + 1);
42         memset(dp,0,sizeof(dp));
43         dp[1] = 1;
44         judge_huiwen();
45         for(int i = 2; i <= len; i++)
46         {
47             dp[i] = dp[i - 1] + 1;
48             for(int j = i - 1; j > 0; j--)
49             {
50                 if(is_h[j][i])
51                 {
52                      dp[i] = min(dp[i], dp[j - 1] + 1);
53                 }
54             }
55         }
56         printf("%d\n",dp[len]);
57     }
58     return 0;
59 }

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