Problem A
Make Palindrome
Input: standard input
Output: standard output
Time Limit: 8 seconds
By definition palindrome is a string which is not changed when reversed. "MADAM" is a nice example of palindrome. It is an easy job to test whether a given string is a palindrome or not. But it may not be so easy to generate a palindrome.
Here we will make a palindrome generator which will take an input string and return a palindrome. You can easily verify that for a string of length 'n', no more than (n-1) characters are required to make it a palindrome. Consider "abcd" and its palindrome "abcdcba" or "abc" and its palindrome "abcba". But life is not so easy for programmers!! We always want optimal cost. And you have to find the minimum number of characters required to make a given string to a palindrome if you are allowed to insert characters at any position of the string.
Input
Each input line consists only of lower case letters. The size of input string will be at most 1000. Input is terminated by EOF.
Output
For each input print the minimum number of characters and such a palindrome seperated by one space in a line. There may be many such palindromes. Any one will be accepted.
Sample Input
abcdaaaaabcaababababaabababapqrsabcdpqrs
Sample Output
3 abcdcba0 aaaa2 abcba1 baab0 abababaabababa9 pqrsabcdpqrqpdcbasrqp
题意：给定字符串。可以在任意位置增添字符，求最少步骤生成回文串。以及生成的串
思路：
和这题一样。http://blog.csdn.net/accelerator_/article/details/11542037。
多开一个vis数组记录状态转移方式便于输出。
代码：
#include <stdio.h>
#include <string.h>
const int MAXN = 1005; char sb[MAXN];
int dp[MAXN][MAXN], vis[MAXN][MAXN], n, i, j; void print(int i, int j) {
if (i == j) {
printf("%c", sb[i]);
return;
}
if (i > j)
return;
if (vis[i][j] == -1) {
printf("%c", sb[i]);
print(i + 1, j - 1);
printf("%c", sb[j]);
}
else if (vis[i][j] == 0) {
printf("%c", sb[i]);
print(i + 1, j);
printf("%c", sb[i]);
}
else if (vis[i][j] == 1) {
printf("%c", sb[j]);
print(i, j - 1);
printf("%c", sb[j]);
}
} int main() {
while (gets(sb) != NULL) {
n = strlen(sb);
for (i = n - 1; i >= 0; i --) {
for (j = i + 1; j < n; j ++) {
if (sb[i] == sb[j]) {
dp[i][j] = dp[i + 1][j - 1];
vis[i][j] = -1;
}
else {
if (dp[i + 1][j] < dp[i][j - 1]) {
dp[i][j] = dp[i + 1][j] + 1;
vis[i][j] = 0;
}
else {
dp[i][j] = dp[i][j - 1] + 1;
vis[i][j] = 1;
}
}
}
}
printf("%d ", dp[0][n - 1]);
print(0, n - 1);
printf("\n");
}
return 0;
}