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)

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <algorithm>
 using namespace std;
 const int MAX =  + ;
 const int INF = 0x3f3f3f3f;
 char s[MAX];
 int dp[MAX],len,is_h[MAX][MAX];
 void judge_huiwen()
 {
     int x,y;
     memset(is_h, , sizeof(is_h));
     for(int i = ; i <= len; i++)
     {
         x = i - ;
         y = i + ;
         while(x >  && y <= len && s[x] == s[y])
         {
             is_h[x][y] = ;
             x--;
             y++;
         }
         x = i;
         y = i + ;
         while(x >  && y <= len && s[x] == s[y])
         {
             is_h[x][y] = ;
             x--;
             y++;
         }
     }
 }
 int main()
 {
     int test;
     scanf("%d", &test);
     while(test--)
     {
         scanf("%s", s + );
         len = strlen(s + );
         memset(dp,,sizeof(dp));
         dp[] = ;
         judge_huiwen();
         for(int i = ; i <= len; i++)
         {
             dp[i] = dp[i - ] + ;
             for(int j = i - ; j > ; j--)
             {
                 if(is_h[j][i])
                 {
                      dp[i] = min(dp[i], dp[j - ] + );
                 }
             }
         }
         printf("%d\n",dp[len]);
     }
     return ;
 }