LightOJ_1025The Specials Menu（区间DP）

最新推荐文章于 2020-11-02 21:38:44 发布

ACMer_wobudong

最新推荐文章于 2020-11-02 21:38:44 发布

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本文链接：https://blog.youkuaiyun.com/ACMer_wobudong/article/details/79452048

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本文介绍了一种通过区间动态规划解决回文串计数问题的方法。该算法用于计算从给定字符串中删除某些字符后形成的所有可能回文串的数量。文章提供了完整的C++代码实现，并通过一个具体的例子说明了如何运用此算法。

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1025 - The Specials Menu

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Time Limit: 2 second(s)

Memory Limit: 32 MB

Feuzem is an unemployed computer scientist who spends his days working at odd-jobs. While on the job he always manages to find algorithmic problems within mundane aspects of everyday life.

Today, while writing down the specials menu at the restaurant he's working at, he felt irritated by the lack of palindromes (strings which stay the same when reversed) on the menu. Feuzem is a big fan of palindromic problems, and started thinking about the number of ways he could remove letters from a particular word so that it would become a palindrome.

Two ways that differ due to order of removing letters are considered the same. And it can also be the case that no letters have to be removed to form a palindrome.

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case contains a single word W (1 ≤ length(W) ≤ 60).

Output

For each case, print the case number and the total number of ways to remove letters from W such that it becomes a palindrome.

Sample Input

Output for Sample Input

SALADS

PASTA

YUMMY

Case 1: 15

Case 2: 8

Case 3: 11

原题连接：http://www.lightoj.com/volume_showproblem.php?problem=1025

题意：求一个字符串删掉若干字母后能组成回文串的个数。

思路：区间DP（DP太菜，完全没想到。。。）dp[l][r]代表 l 到 r 区间内回文串的个数，

转移方程dp[l][r]=dp[l+1][r]+dp[l][r-1]。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<set>
#include<map>
#include<string>
#include<queue>
#include<stack>
#include<sstream>
#include<ctype.h>
#include<vector>
#include<list>
using namespace std;
typedef long long ll;
const int INF=1<<30;
const double eps=0.0000001;
const int maxn = 60+5;
ll dp[maxn][maxn];           //dp[l][r]代表区间l到r之间的回文串个数

int main()
{
    int T;
    scanf("%d",&T);
    int cas=1;
    while(T--)
    {
        char s[100];
        memset(dp,0,sizeof(dp));
        scanf("%s",s+1);
        int len=strlen(s+1);
        for(int i=1;i<=len;i++)
        {
            for(int l=1;l+i-1<=len;l++)
            {
                int r=l+i-1;
                dp[l][r]+=dp[l+1][r];
                dp[l][r]+=dp[l][r-1];
                if(s[l]==s[r])
                    dp[l][r]++;
                else
                    dp[l][r]-=dp[l+1][r-1];
            }
        }
        printf("Case %d: %lld\n",cas++,dp[1][len]);
    }
    return 0;
}