UVa11584Partitioning by Palindromes(字符串区间dp)

UVa11584Partitioning by Palindromes(字符串区间dp)

题意：给定一个字符串s, 问说最少可以划分成几个回文串。

思路：dp[i]表示从1到第i个字符最少可以划分为几个回文，状态转移方程

dp[i] = min(dp[i], dp[j-1]+1), 如果满足 s[j] 到 s[i] 为回文字符串。

用 judge 函数判断从j到i是否可以形成回文串。

Sample Input

3

racecar

fastcar

aaadbccb

Sample Output

1

7

3

<span style="font-size:18px;">#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;

const double PI = acos(-1.0);
const double e = 2.718281828459;
const double eps = 1e-8;
const int MAXN = 1010;
char s[MAXN];
int dp[MAXN];

int judge(int l, int r)
{
while(l <= r)
{
if(s[l] != s[r])
return 0;
l++;
r--;
}
return 1;
}

int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
int Case;
cin>>Case;
while(Case--)
{
scanf("%s", s+1);
int len = strlen(s+1);
dp[0] = 0;
for(int i = 1; i <= len; i++)
dp[i] = 1010;
for(int i = 1; i <= len; i++)
{
for(int j = 1; j <= i; j++)
{
if(judge(j, i))
dp[i] = min(dp[i], dp[j-1]+1);
}
}
cout<<dp[len]<<endl;
}
return 0;
}</span>