UVA 11584 Partitioning by Palindromes

Partitioning by Palindromes

Time Limit: 1000ms

Memory Limit: 131072KB

This problem will be judged on UVA. Original ID: 11584
64-bit integer IO format: %lld      Java class name: Main

 

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。。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 1010;
char str[maxn];
int dp[maxn];
bool check(int i,int j){
    for(int a = i,b = j; a < b; ++a,--b)
        if(str[a] != str[b]) return false;
    return true;
}
int main() {
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%s",str+1);
        int len = strlen(str+1);
        for(int i = 0; i < maxn; ++i) dp[i] = INF;
        dp[0] = 0;
        for(int i = 1; i <= len; ++i){
            for(int j = 1; j <= i; ++j)
                if(check(j,i)) dp[i] = min(dp[i],dp[j-1]+1);
        }
        printf("%d
",dp[len]);
    }
    return 0;
}