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：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <set>
 #include <map>
 #include <list>
 #include <unordered_set>
 #include <unordered_map>
 #include <cmath>

 #define FRER() freopen("in.txt", "r", stdin)
 #define FREW() freopen("out.txt", "w", stdout)
 #define INF 0x3f3f3f3f

 using namespace std;

 const int maxn =  + ;

 int n, len, l, r;
 char str[maxn];

 bool check[maxn][maxn];
 int dp[maxn];

 int main()
 {
     //FRER();
     scanf("%d", &n);
     while(n--) {
         scanf("%s", str);
         len = strlen(str);
         memset(check, , sizeof(check));

         for(int i = ; i < len; ++i) {
             l = r = i;
             while(l >=  && r < len) {
                 if(str[l] == str[r]) {
                     check[l + ][r + ] = ;
                     --l; ++r;
                 }
                 else break;
             }
             l = i;
             r = i + ;

             while(l >=  && r < len) {
                 if(str[l] == str[r]) {
                     check[l + ][r + ] = ;
                     --l; ++r;
                 }
                 else break;
             }
         }

         memset(dp, , sizeof(dp));
         for(int i = ; i <= len; ++i) {
             dp[i] = dp[i - ] + ;
             for(int j = ; j < i; ++j) {
                 if(check[j][i])
                     dp[i] = min(dp[i], dp[j - ] + );
             }
         }

         printf("%d\n", dp[len]);
     }
     return ;
 }