lightoj 1025 - The Specials Menu 【区间dp】

题目链接：lightoj 1025 - The Specials Menu

1025 - The Specials Menu
PDF (English) Statistics Forum
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：求区间回文子序列。。。

思路：明显的区间dp。
dp[i][j]表示区间[i, j]的答案。转移有：
if(str[i]!=str[j])：dp[i][j]=dp[i+1][j]+dp[i][j−1]−dp[i+1][j−1]
else:dp[i][j]=dp[i+1][j]+dp[i][j−1]+1

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#include <vector>#include <map>#define ll o<<1#define rr o<<1|1#define fi first#define se second#define CLR(a, b) memset(a, (b), sizeof(a))using namespace std;typedef long long LL;typedef pair<int, int> pii;const int INF = 0x3f3f3f3f;const int MAXN = 3*1e5 + 10;LL dp[100][100];char str[100];LL DFS(int i, int j) {    if(i > j) return dp[i][j] = 0;    if(dp[i][j] != -1) return dp[i][j];    if(i == j) return dp[i][j] = 1;    LL ans = DFS(i+1, j) + DFS(i, j-1);    if(str[i] == str[j]) ans += 1;    else ans -= DFS(i+1, j-1);    dp[i][j] = ans;    return ans;}int main(){    int t, kcase = 1; scanf("%d", &t);    while(t--) {        scanf("%s", str+1);        int len = strlen(str+1); CLR(dp, -1);        printf("Case %d: %lld\n", kcase++, DFS(1, len));    }    return 0;}