UVA 11584 Partitioning by Palindromes

We say a sequence of characters is a palindrome if it is the same written forwards and backwards. For example, ‘racecar‘ is a palindrome, but ‘fastcar‘ is not.

A partition of a sequence of characters is a list of one or more disjoint non-empty groups of consecutive characters whose concatenation yields the initial sequence. For example, (‘race‘, ‘car‘) is a partition of ‘racecar‘ into two groups.

Given a sequence of characters, we can always create a partition of these characters such that each group in the partition is a palindrome! Given this observation it is natural to ask: what is the minimum number of groups needed for a given string such that every group is a palindrome?

For example:

• ‘racecar‘ is already a palindrome, therefore it can be partitioned into one group.
• ‘fastcar‘ does not contain any non-trivial palindromes, so it must be partitioned as (‘f‘, ‘a‘, ‘s‘, ‘t‘, ‘c‘, ‘a‘, ‘r‘).
• ‘aaadbccb‘ can be partitioned as (‘aaa‘, ‘d‘, ‘bccb‘).

Input begins with the number n of test cases. Each test case consists of a single line of between 1 and 1000 lowercase letters, with no whitespace within.

For each test case, output a line containing the minimum number of groups required to partition the input into groups of palindromes.

Sample Input

3
racecar
fastcar
aaadbccb

Sample Output

1
7
3

解题：dp,求一个串最少的回文串数。。。。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 1010;
18 char str[maxn];
19 int dp[maxn];
20 bool check(int i,int j){
21     for(int a = i,b = j; a < b; ++a,--b)
22         if(str[a] != str[b]) return false;
23     return true;
24 }
25 int main() {
26     int t;
27     scanf("%d",&t);
28     while(t--){
29         scanf("%s",str+1);
30         int len = strlen(str+1);
31         for(int i = 0; i < maxn; ++i) dp[i] = INF;
32         dp[0] = 0;
33         for(int i = 1; i <= len; ++i){
34             for(int j = 1; j <= i; ++j)
35                 if(check(j,i)) dp[i] = min(dp[i],dp[j-1]+1);
36         }
37         printf("%d\n",dp[len]);
38     }
39     return 0;
40 }

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