LPS UVA 11404 Palindromic Subsequence

题目传送门

题意：求LPS (Longest Palidromic Subsequence) 最长回文子序列。和回文串不同，子序列是可以不连续的。

分析：1. 推荐->还有一种写法是用了LCS的思想，dp[i][j]表示i到j的最长回文串长度，状态转移方程：

　　　　　　1. dp[j][j+i-1] = dp[j+1][j+i-2] + 2; (str[j] == str[j+i-1])

　　　　　　2. dp[j][j+i-1] = max (dp[j+1][j+i-1], dp[j][j+i-2]); (str[j] != str[j+i-1])

　　　2. 转化为LCS问题，将字符串逆序，然后和本串求LCS就是LPS的长度，但是前一半是LPS的一半，可以补全

代码1：

/************************************************
 * Author        :Running_Time
 * Created Time  :2015-8-7 14:26:22
 * File Name     :UVA_11404.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
string ans[MAXN][MAXN];
int dp[MAXN][MAXN];
char str[MAXN];

void LPS(void)  {
    int len = strlen (str + 1);
    memset (dp, 0, sizeof (dp));
    for (int i=1; i<=len; ++i)  dp[i][i] = 1;
    for (int i=1; i<=len; ++i)  ans[i][i] = str[i];

    for (int i=2; i<=len; ++i)   {              //区间长度
        for (int j=1; j+i-1<=len; ++j)  {       //[j, j+i-1]
            if (str[j] == str[j+i-1])   {
                if (i == 2) {
                    dp[j][j+i-1] = 2;
                    ans[j][j+i-1] = ans[j][j] + ans[j+i-1][j+i-1];  continue;
                }
                dp[j][j+i-1] = dp[j+1][j+i-2] + 2;
                ans[j][j+i-1] = str[j] + ans[j+1][j+i-2] + str[j+i-1];
            }
            else if (dp[j+1][j+i-1] > dp[j][j+i-2]) {
                dp[j][j+i-1] = dp[j+1][j+i-1];
                ans[j][j+i-1] = ans[j+1][j+i-1];
            }
            else if (dp[j][j+i-2] > dp[j+1][j+i-1]) {
                dp[j][j+i-1] = dp[j][j+i-2];
                ans[j][j+i-1] = ans[j][j+i-2];
            }
            else    {
                dp[j][j+i-1] = dp[j+1][j+i-1];
                ans[j][j+i-1] = min (ans[j+1][j+i-1], ans[j][j+i-2]);
            }
        }
    }
    int mlen = dp[1][len];
    for (int i=0; i<mlen; ++i)  {
        printf ("%c", ans[1][len][i]);
    }
    puts ("");
}

int main(void)    {     //UVA 11404 Palindromic Subsequence
    while (scanf ("%s", str + 1) == 1)  {
        LPS ();
    }

    return 0;
}

代码2：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-7 14:26:22
* File Name     :UVA_11404.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct Ans  {
    int len;
    string s;
}dp[MAXN][MAXN];
char str[MAXN], rstr[MAXN];

int main(void)    {
    while (scanf ("%s", str + 1) == 1)  {
        int len = strlen (str + 1);
        for (int i=1; i<=len; ++i)  {
            rstr[len-i+1] = str[i];
        }
        for (int i=0; i<=len; ++i)  {
            dp[0][i].len = 0, dp[0][i].s = "";
        }
        for (int i=1; i<=len; ++i)  {
            for (int j=1; j<=len; ++j)  {
                if (str[i] == rstr[j])  {
                    dp[i][j].len = dp[i-1][j-1].len + 1;
                    dp[i][j].s = dp[i-1][j-1].s + str[i];
                }
                else if (dp[i][j-1].len > dp[i-1][j].len)   {
                    dp[i][j].len = dp[i][j-1].len;
                    dp[i][j].s = dp[i][j-1].s;
                }
                else if (dp[i-1][j].len > dp[i][j-1].len)   {
                    dp[i][j].len = dp[i-1][j].len;
                    dp[i][j].s = dp[i-1][j].s;
                }
                else    {
                    dp[i][j].len = dp[i-1][j].len;
                    dp[i][j].s = min (dp[i-1][j].s, dp[i][j-1].s);
                }
            }
        }
        int mlen = dp[len][len].len;
        string ans = dp[len][len].s;
        for (int i=0; i<mlen/2; ++i)  printf ("%c", ans[i]);
        int j;
        if (mlen & 1)   j = mlen / 2;
        else    j = mlen / 2 - 1;
        for (; j>=0; --j)   printf ("%c", ans[j]);
        puts ("");
    }

    return 0;
}