lightoj1044--Palindrome Partitioning(记忆化搜索)

Description

A palindrome partition is the partitioning of a string such that each separate substring is a palindrome.

For example, the string "ABACABA" could be partitioned in several different ways, such as {"A","B","A","C","A","B","A"}, {"A","BACAB","A"}, {"ABA","C","ABA"}, or {"ABACABA"}, among others.

You are given a string s. Return the minimum possible number of substrings in a palindrome partition of s.

Input

Input starts with an integer T (≤ 40), denoting the number of test cases.

Each case begins with a non-empty string s of uppercase letters with length no more than 1000.

Output

For each case of input you have to print the case number and the desired result.

Sample Input

3

AAAA

ABCDEFGH

QWERTYTREWQWERT

Sample Output

Case 1: 1

Case 2: 8

Case 3: 5

题意：给你一个字符串，让你找出回文串的最少个数

思路：这是一道动态规划，将一条字符串切割成两条（或者不切割），判断两条子串符合条件的值的和的最小值

优化思路：先给字符串的回文子串打表ok[i][j]，在切割的同时判断右半边是否是回文的，是则d[i]=min(d[i],1+d[i-j]),不是说明这种情况必定不是最优

递归代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <cmath>#include <stack>#include <map>using namespace std;typedef long long ll;string a;int d[1005][1005];int ok[1005][1005];int dp(int l,int r){    if(d[l][r]>=0)return d[l][r];    if(ok[l][r])return d[l][r]=1;    int minn=1001;    for(int i=l; i<r; i++)        if(ok[l][i])        minn=min(minn,1+dp(i+1,r));    return d[l][r]=minn;}int main(){    int t,fir=1;    cin>>t;    while(t--)    {        cin>>a;        memset(d,-1,sizeof(d));        memset(ok,0,sizeof(ok));        int n=a.length();        for(int i=0; i<n; i++)ok[i][i]=1;        for(int i=0; i<n-1; i++)ok[i][i+1]=(a[i]==a[i+1]);        for(int len=3; len<=n; len++)            for(int i=0; i+len-1<n; i++)            {                int j=i+len-1;                ok[i][j]=(a[i]==a[j])&&ok[i+1][j-1];            }        printf("Case %d: %d\n",fir++,dp(0,n-1));    }    return 0;}

非递归代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <cmath>#include <stack>#include <map>using namespace std;typedef long long ll;string a;int d[1005];int ok[1005][1005];int main(){    int t,fir=1;    cin>>t;    while(t--)    {        cin>>a;        memset(ok,0,sizeof(ok));        int n=a.length();        for(int i=0; i<n; i++)ok[i][i]=1;        for(int i=0; i<n-1; i++)ok[i][i+1]=(a[i]==a[i+1]);        for(int len=3; len<=n; len++)            for(int i=0; i+len-1<n; i++)            {                int j=i+len-1;                ok[i][j]=(a[i]==a[j])&&ok[i+1][j-1];            }        for(int i=0; i<n; i++)        {            d[i]=i+1;            for(int j=0; j<=i; j++)                if(ok[j][i])                {                    if(j==0)                        d[i]=1;                    else                        d[i]=min(d[i],1+d[j-1]);                }        }        printf("Case %d: %d\n",fir++,d[n-1]);    }    return 0;}

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