LightOJ 1025 - The Specials Menu （区间dp）

题意:

求字符串删去一些字符,使得剩下的串是回文串的方法,不删也可以

分析:

区间dp,我们可以换一个问法,求给定字符串的回文子序列有多少
dp[l][r]:=[l,r]区间删去一些字符,使得剩下的字符串是回文串的方法数
转移参考最长回文子序列的
dp[l][r]=dp[l+1][r]+dp[l][r+1]−dp[l+1][r−1],删s[l],s[r],多算了都删的
s[l]==s[r],dp[l][r]+=dp[l+1][r−1]+1,删s[l]和s[r],或者删去s[l+1⋯r−1]

代码:

////  Created by TaoSama on 2015-11-02//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n;char s[65];typedef long long LL;LL dp[65][65];int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        scanf("%s", s + 1);        n = strlen(s + 1);        memset(dp, 0, sizeof dp);        for(int len = 1; len <= n; ++len) {            for(int i = 1; i + len - 1 <= n; ++i) {                int j = i + len - 1;                dp[i][j] += dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1];                if(s[i] == s[j]) dp[i][j] += dp[i + 1][j - 1] + 1;            }        }        printf("Case %d: %lld\n", ++kase, dp[1][n]);    }    return 0;}