LPS UVA 11404 Palindromic Subsequence

题目传送门

 1 /*
 2     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。
 3         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(为啥不就是LPS?)，但是前一半是LPS的一半，可以补全
 4         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：
 5             1. dp[j][j+i-1] = dp[j+1][j+i-2] + 2;   (str[j] == str[j+i-1])
 6             2. dp[j][j+i-1] = max (dp[j+1][j+i-1], dp[j][j+i-2]);   (str[j] != str[j+i-1])
 7 */
 8 /************************************************
 9  * Author        :Running_Time
10  * Created Time  :2015-8-7 14:26:22
11  * File Name     :UVA_11404.cpp
12  ************************************************/
13
14 #include <cstdio>
15 #include <algorithm>
16 #include <iostream>
17 #include <sstream>
18 #include <cstring>
19 #include <cmath>
20 #include <string>
21 #include <vector>
22 #include <queue>
23 #include <deque>
24 #include <stack>
25 #include <list>
26 #include <map>
27 #include <set>
28 #include <bitset>
29 #include <cstdlib>
30 #include <ctime>
31 using namespace std;
32
33 #define lson l, mid, rt << 1
34 #define rson mid + 1, r, rt << 1 | 1
35 typedef long long ll;
36 const int MAXN = 1e3 + 10;
37 const int INF = 0x3f3f3f3f;
38 const int MOD = 1e9 + 7;
39 string ans[MAXN][MAXN];
40 int dp[MAXN][MAXN];
41 char str[MAXN];
42
43 int main(void)    {     //UVA 11404 Palindromic Subsequence
44     while (scanf ("%s", str + 1) == 1)  {
45         int len = strlen (str + 1);
46         memset (dp, 0, sizeof (dp));
47         for (int i=1; i<=len; ++i)  dp[i][i] = 1;
48         for (int i=1; i<=len; ++i)  ans[i][i] = str[i];
49
50         for (int i=2; i<=len; ++i)   {      //区间长度
51             for (int j=1; j+i-1<=len; ++j)  {       //[j, j+i-1]
52                 if (str[j] == str[j+i-1])   {
53                     if (i == 2) {
54                         dp[j][j+i-1] = 2;
55                         ans[j][j+i-1] = ans[j][j] + ans[j+i-1][j+i-1];  continue;
56                     }
57                     dp[j][j+i-1] = dp[j+1][j+i-2] + 2;
58                     ans[j][j+i-1] = str[j] + ans[j+1][j+i-2] + str[j+i-1];
59                 }
60                 else if (dp[j+1][j+i-1] > dp[j][j+i-2]) {
61                     dp[j][j+i-1] = dp[j+1][j+i-1];
62                     ans[j][j+i-1] = ans[j+1][j+i-1];
63                 }
64                 else if (dp[j][j+i-2] > dp[j+1][j+i-1]) {
65                     dp[j][j+i-1] = dp[j][j+i-2];
66                     ans[j][j+i-1] = ans[j][j+i-2];
67                 }
68                 else    {
69                     dp[j][j+i-1] = dp[j+1][j+i-1];
70                     ans[j][j+i-1] = min (ans[j+1][j+i-1], ans[j][j+i-2]);
71                 }
72             }
73         }
74         int mlen = dp[1][len];
75         for (int i=0; i<mlen; ++i)  {
76             printf ("%c", ans[1][len][i]);
77         }
78         puts ("");
79     }
80
81     return 0;
82 }

 1 /************************************************
 2 * Author        :Running_Time
 3 * Created Time  :2015-8-7 14:26:22
 4 * File Name     :UVA_11404.cpp
 5  ************************************************/
 6
 7 #include <cstdio>
 8 #include <algorithm>
 9 #include <iostream>
10 #include <sstream>
11 #include <cstring>
12 #include <cmath>
13 #include <string>
14 #include <vector>
15 #include <queue>
16 #include <deque>
17 #include <stack>
18 #include <list>
19 #include <map>
20 #include <set>
21 #include <bitset>
22 #include <cstdlib>
23 #include <ctime>
24 using namespace std;
25
26 #define lson l, mid, rt << 1
27 #define rson mid + 1, r, rt << 1 | 1
28 typedef long long ll;
29 const int MAXN = 1e3 + 10;
30 const int INF = 0x3f3f3f3f;
31 const int MOD = 1e9 + 7;
32 struct Ans  {
33     int len;
34     string s;
35 }dp[MAXN][MAXN];
36 char str[MAXN], rstr[MAXN];
37
38 int main(void)    {
39     while (scanf ("%s", str + 1) == 1)  {
40         int len = strlen (str + 1);
41         for (int i=1; i<=len; ++i)  {
42             rstr[len-i+1] = str[i];
43         }
44         for (int i=0; i<=len; ++i)  {
45             dp[0][i].len = 0, dp[0][i].s = "";
46         }
47         for (int i=1; i<=len; ++i)  {
48             for (int j=1; j<=len; ++j)  {
49                 if (str[i] == rstr[j])  {
50                     dp[i][j].len = dp[i-1][j-1].len + 1;
51                     dp[i][j].s = dp[i-1][j-1].s + str[i];
52                 }
53                 else if (dp[i][j-1].len > dp[i-1][j].len)   {
54                     dp[i][j].len = dp[i][j-1].len;
55                     dp[i][j].s = dp[i][j-1].s;
56                 }
57                 else if (dp[i-1][j].len > dp[i][j-1].len)   {
58                     dp[i][j].len = dp[i-1][j].len;
59                     dp[i][j].s = dp[i-1][j].s;
60                 }
61                 else    {
62                     dp[i][j].len = dp[i-1][j].len;
63                     dp[i][j].s = min (dp[i-1][j].s, dp[i][j-1].s);
64                 }
65             }
66         }
67         int mlen = dp[len][len].len;
68         string ans = dp[len][len].s;
69         for (int i=0; i<mlen/2; ++i)  printf ("%c", ans[i]);
70         int j;
71         if (mlen & 1)   j = mlen / 2;
72         else    j = mlen / 2 - 1;
73         for (; j>=0; --j)   printf ("%c", ans[j]);
74         puts ("");
75     }
76
77     return 0;
78 }

LCS