UVa 11404 Palindromic Subsequence (LCS)

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;
}