Light oj 1025 （区间dp）

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
3 SALADS PASTA YUMMY	Case 1: 15 Case 2: 8 Case 3: 11

PROBLEM SETTER: MUNTASIR MUZAHID CHOWDHURY
SPECIAL THANKS: JANE ALAM JAN (DATASET)

题意：删除字母使得串为回文串的方案数

/*
题目链接: http://lightoj.com/volume_showproblem.php?problem=1025 
*/

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-8

typedef long long ll;

using namespace std;

#define N 65

ll dp

;
char c
;

int main()
{
    int i,j,t,ca=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",c);
        int len=strlen(c);
        memset(dp,0,sizeof(dp));
        for(int i=0;i<len;i++)
            dp[i][i]=1;

        for(int i=len-1;i>=0;i--)
            for(int j=i+1;j<len;j++)
            {
               dp[i][j]=dp[i][i]+dp[i+1][j];
               for(int k=i+1;k<=j;k++)
                  if(c[i]==c[k]) dp[i][j]+=dp[i+1][k-1]+1;
            }
        printf("Case %d: %lld\n",++ca,dp[0][len-1]);
    }
   return 0;
}