标签：style blog http color io os ar java for
A Subsequence is a sequence obtained by deleting zero or more characters in a string. A Palindrome is a string which when read from left to right, reads same as when read from right to left. Given a string, find the longest palindromic subsequence. If there are many answers to it, print the one that comes lexicographically earliest.
Constraints
Input consists of several strings, each in a separate line. Input is terminated by EOF.
For each line in the input, print the output in a single line.
aabbaabb computer abzla samhita
aabbaa c aba aha
解题：求最长的且字典序最小的回文子序列。把原串逆转，然后与原串求LCS。LCS的的前半部分一定要求的回文序列的前半部分，但是后半部分可能不是。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib> 10 #include <string> 11 #include <set> 12 #include <stack> 13 #define LL long long 14 #define pii pair<int,int> 15 #define INF 0x3f3f3f3f 16 using namespace std; 17 const int maxn = 1010; 18 struct DP{ 19 int len; 20 string str; 21 }; 22 DP dp[maxn][maxn]; 23 char sa[maxn],sb[maxn]; 24 int main() { 25 while(gets(sa)){ 26 int len = strlen(sa); 27 strcpy(sb,sa); 28 reverse(sb,sb+len); 29 for(int i = 0; i <= len; ++i){ 30 dp[0][i].len = 0; 31 dp[0][i].str = ""; 32 } 33 for(int i = 1; i <= len; ++i){ 34 for(int j = 1; j <= len; ++j){ 35 if(sa[i-1] == sb[j-1]){ 36 dp[i][j].len = dp[i-1][j-1].len+1; 37 dp[i][j].str = dp[i-1][j-1].str + sa[i-1]; 38 }else if(dp[i-1][j].len > dp[i][j-1].len){ 39 dp[i][j].len = dp[i-1][j].len; 40 dp[i][j].str = dp[i-1][j].str; 41 }else if(dp[i-1][j].len < dp[i][j-1].len){ 42 dp[i][j].len = dp[i][j-1].len; 43 dp[i][j].str = dp[i][j-1].str; 44 }else{ 45 dp[i][j].len = dp[i-1][j].len; 46 dp[i][j].str = min(dp[i-1][j].str,dp[i][j-1].str); 47 } 48 } 49 } 50 string ans = dp[len][len].str; 51 if(dp[len][len].len&1){ 52 for(int i = 0; i < dp[len][len].len>>1; ++i) 53 putchar(ans[i]); 54 for(int i = dp[len][len].len>>1; i >= 0; --i) 55 putchar(ans[i]); 56 putchar(‘\n‘); 57 }else{ 58 for(int i = 0; i+1 < dp[len][len].len>>1; ++i) 59 putchar(ans[i]); 60 for(int i = dp[len][len].len>>1; i >= 0; --i) 61 putchar(ans[i]); 62 putchar(‘\n‘); 63 } 64 } 65 return 0; 66 }
UVA 11404 Palindromic Subsequence
标签：style blog http color io os ar java for
原文地址：http://www.cnblogs.com/crackpotisback/p/4037278.html
