例题9-7 划分成回文串 UVa11584

1.题目描述：点击打开链接

2.解题思路：本题要求划分回文串，且个数尽可能的少。可以用动态规划解决。先提前判断i~j是否构成回文串，时间复杂度是O(N^2)，然后定义d(i)表示0~i-1划分成的回文串的最小个数。则状态转移方程为：

d(i)=min(d(i),d(j)+1)(s[j...i]是回文串)

上式中，d(i)的初始值是i,这样每次判断只需要O(1)的时间，总时间复杂度是O(N^2)。当然，判断回文串的过程可以和状态转移相结合，细节请参考第二份代码。

3.代码：

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<algorithm>
#include<string>
#include<sstream>
#include<set>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<deque>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<functional>
using namespace std;

#define maxn 1030000
int vis[maxn];
int d[maxn];
int n;
int main()
{
	//freopen("test.txt", "r", stdin);
	cin >> n;
	while (n--)
	{
		string str;
		cin >> str;
		memset(vis, 0, sizeof(vis));
		memset(d, 0, sizeof(d));
		int len = str.length();
		for (int i = 0; i <= len; i++)d[i] = i;
		for (int i = 0; i < 2 * len - 1; i++)//枚举中心，判断i到j是否为回文串
		{
			int p;
			int m = i / 2;
			for (int j = m; j >=0; j--)
			{
				if (i - j<len&&str[j] == str[i - j])
				{
					p = (j << 10) | (i - j);
					vis[p] = 1;
				}
				else break;
			}
		}
		int q = 0;
		for (int i = 0; i <= len; i++)//枚举起点，终点
		{
			for (int j = 0; j <= i - 1; j++)
			{
				q = (j << 10) | (i - 1);
				if (vis[q])
					d[i] = min(d[i], d[j] + 1);
			}
		}
		printf("%d\n", d[len]);
	}
	return 0;
}

参考代码：

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<algorithm>
#include<string>
#include<sstream>
#include<set>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<deque>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<functional>
using namespace std;

#define N 1002

int f[N];
bool d[N][N];

int main()
{
	freopen("test.txt","r", stdin);
	int n, t, i, j, result;
	cin >> t;
	while (t--)
	{
		string s;
		cin >> s;
		n = s.length();
		memset(f, 1000000, sizeof(f));
		memset(d, false, sizeof(d));
		for (int i = 0; i <= n; i++)//f[i]表示i到n划分成的最小回文串的个数
			f[i] = n - i;
		for (int i = n - 1; i >= 0; i--)//i是起点，逆序枚举
		for (int j = i; j < n; j++)//j是终点，顺序枚举
		if (s[i] == s[j] && (j - i < 2 || d[i + 1][j - 1])) //如果j-i<2那么只需要直接用s[i]==s[j]判断即可，否则，用d[i+1][j-1]来判断是否为回文串
		{
			d[i][j] = true;
			f[i] = min(f[i], f[j + 1] + 1);
		}
		cout << f[0] << endl;
	}
	return 0;
}