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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 }
LPS UVA 11404 Palindromic Subsequence
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原文地址：http://www.cnblogs.com/Running-Time/p/4711165.html
