given a word,convert it into a pallindrome with minimum addition of letters to it.letters can be added anywhere in the word.for eg if yahoo is given result shud be yahoohay.give a optimize soln
方法1：
設字符串str1, 其倒置字符串str2, 求得str1和str2的最長公共序列str3
對str1,str2,str3做三路歸并
方法2：
dp[i][dis]表示從i開始的長度為dis的字符串，要變為回文，最少添加多少字符
#include <iostream>
#include <stdio.h>using namespace std;const int N = 1000;
char ans[N];
int dp[N][N];
char s[N];int tracebace(int p, int dis, int i, bool &flag) {if (dis == 0) {flag = false;return 0;}if (dis == 1) {flag = true;ans[i] = s[p];return 1;}if (s[p] == s[p+dis-1]) {ans[i] = s[p];return 1 + tracebace(p + 1, dis -2, i +1, flag);}if (dp[p][dis] == dp[p+1][dis-1] + 1) {ans[i] = s[p];return 1 + tracebace(p + 1, dis - 1, i + 1, flag);}else {ans[i] = s[p+dis-1];return 1 + tracebace(p, dis - 1, i + 1, flag);}
}int main() {scanf("%s", s);memset(dp,0,sizeof(dp));int len = strlen(s);for (int dis = 2; dis <= len; ++dis ) {for (int i = 0; i <= len - dis; ++i) {if (s[i] == s[i + dis - 1])dp[i][dis] = dp[i + 1][dis - 2];else {dp[i][dis] = 1 + min(dp[i+1][dis-1], dp[i][dis-1]);}}}bool flag = true;int preLen = tracebace(0, len, 0, flag);int dis = flag ? 1 : 0;for (int i = preLen; i < len + dp[0][len]; ++i,++dis) {ans[i] = ans[preLen-dis-1];}ans[len + dp[0][len]] = 0;cout << ans<<endl;return 0;
}
