UVa 11404 - Palindromic Subsequence （最长回文子序列　ＤＰ）

Palindromic Subsequence

Time Limit:3000MS

Memory Limit:Unknown

64bit IO Format:%lld & %llu

[Submit] [Go Back] [Status]

Description

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

Maximum length of string is 1000.
Each string has characters `a' to `z' only.

Input

Input consists of several strings, each in a separate line. Input is terminated byEOF.

Output

For each line in the input, print the output in a single line.

Sample Input

aabbaabb
computer
abzla
samhita

Sample Output

aabbaa
c
aba
aha

题意：
给定一个由小写字母组成的字符串，删除其中的0个或多个字符，使得剩下的（顺序不变）组成一个尽量长的回文串。如果有多解，输出字典序最小的。

思路：

可以转化为ＬＣＳ问题，就是该字符串与它逆序串的ＬＣＳ

但是求出来的ＬＣＳ不一定是回文串，比如：

kfclbckibbibjccbej

LCS : bcibbibc

结果应该是：bcibbicb

因此由它前len/2可得到回文串

第二种见下面

因为长度num 定义为了char，导致无限ＷＡ

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;

//#define WIN
#ifdef WIN
typedef __int64 LL;
#define iform "%I64d"
#define oform "%I64d\n"
#define oform1 "%I64d"
#else
typedef long long LL;
#define iform "%lld"
#define oform "%lld\n"
#define oform1 "%lld"
#endif

#define S64I(a) scanf(iform, &(a))
#define P64I(a) printf(oform, (a))
#define P64I1(a) printf(oform1, (a))
#define REP(i, n) for(int (i)=0; (i)<n; (i)++)
#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)
#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)

const int INF = 0x3f3f3f3f;
const double eps = 10e-9;
const double PI = (4.0*atan(1.0));

const int maxn = 1000 + 20;
char s1[maxn];
char s2[maxn];
int dp[maxn][maxn];
string ansA[maxn][maxn];
int n;

void LCS() {
for(int i=0; i<=n; i++) {
dp[0][i] = 0;
ansA[0][i] = "";
}
for(int i=1; i<=n; i++) {
for(int j=1; j<=n; j++) {
dp[i][j] = 0;
ansA[i][j] = "";
if(s1[i] == s2[j]) {
dp[i][j] = dp[i-1][j-1] + 2;
ansA[i][j] = ansA[i-1][j-1] + s1[i];
}
if(dp[i-1][j] > dp[i][j] || (dp[i-1][j] == dp[i][j] && ansA[i-1][j] < ansA[i][j])) {
dp[i][j] = dp[i-1][j];
ansA[i][j] = ansA[i-1][j];
}
if(dp[i][j-1] > dp[i][j] || (dp[i][j-1] == dp[i][j] && ansA[i][j-1] < ansA[i][j])) {
dp[i][j] = dp[i][j-1];
ansA[i][j] = ansA[i][j-1];
}
}
}
}

char ans[maxn];

int main() {

while(scanf("%s", s1+1) != EOF) {
n = strlen(s1+1);
for(int i=1; i<=n; i++) s2[i] = s1[n-i+1];
s2[n+1] = '\0';
LCS();
string aans = ansA

;
strcpy(ans, aans.c_str());
int num = strlen(ans);
for(int i=0; i<num/2; i++) {
ans[num-i-1] = ans[i];
}
puts(ans);
}

return 0;
}

再贴上另外一种思路：

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;

//#define WIN
#ifdef WIN
typedef __int64 LL;
#define iform "%I64d"
#define oform "%I64d\n"
#define oform1 "%I64d"
#else
typedef long long LL;
#define iform "%lld"
#define oform "%lld\n"
#define oform1 "%lld"
#endif

#define S64I(a) scanf(iform, &(a))
#define P64I(a) printf(oform, (a))
#define P64I1(a) printf(oform1, (a))
#define REP(i, n) for(int (i)=0; (i)<n; (i)++)
#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)
#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)

const int INF = 0x3f3f3f3f;
const double eps = 10e-9;
const double PI = (4.0*atan(1.0));

const int maxn = 1000 + 20;
char str[maxn];
int dp[maxn][maxn];
string ansA[maxn][maxn];

int f(int st, int ed) {
if(ed < st) return 0;
if(st == ed) {
ansA[st][ed] = str[st];
return 1;
}
if(dp[st][ed] != -1) return dp[st][ed];
int ret = 0;
string rets = "";
if(str[st] == str[ed]) {
ret = f(st+1, ed-1) + 2;
rets = str[st] + ansA[st+1][ed-1] + str[ed];
}
int t = f(st+1, ed);
if(t > ret || (t == ret && ansA[st+1][ed] < rets)) {
ret = t;
rets = ansA[st+1][ed];
}
t = f(st, ed-1);
if(t > ret || (t == ret && ansA[st][ed-1] < rets)) {
ret = t;
rets = ansA[st][ed-1];
}
ansA[st][ed] = rets;
return dp[st][ed] = ret;
}

int main() {

while(scanf("%s", str) != EOF) {
int n = strlen(str);
memset(dp, -1, sizeof(dp));
int ans = f(0, n-1);
puts(ansA[0][n-1].c_str());
}

return 0;
}