UVa 11404 Palindromic Subsequence (LCS)

时间：2022-08-03 04:47:43

题意：给定一个字符串，问删除一些字符，使得它成为一个最长回访串，如果有多个，输出字典序最小的那个。

析： 我们可以把原字符串反转，然后求两个串的LCS，就得到最长回文串，不过要注意一些细节。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <unordered_map>

#include <unordered_set>

#define debug() puts("++++");

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1000 + 5;

const int mod = 2000;

const int dr[] = {-1, 1, 0, 0};

const int dc[] = {0, 0, 1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

struct Node{

    int val;

    string str;

    bool operator > (const Node &p) const{

        if(val != p.val)  return val > p.val;

        return str < p.str;

    }

};

Node dp[maxn][maxn];

char s[maxn];

int main(){

    ios::sync_with_stdio(false);

    while(cin >> s+1){

        n = strlen(s+1);

        for(int i = n; i > 0; --i){

            dp[i][i].val = 1;

            dp[i][i].str = s[i];

            for(int j = i+1; j <= n; ++j)

                if(s[i] == s[j])  dp[i][j].val = dp[i+1][j-1].val + 2, dp[i][j].str = s[i] + dp[i+1][j-1].str + s[j];

                else if(dp[i+1][j] > dp[i][j-1])  dp[i][j] = dp[i+1][j];

                else  dp[i][j] = dp[i][j-1];

        }

        cout << dp[1][n].str << endl;

    }

    return 0;

}

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