usaco 1.32

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1.求最长回文子序列：我的想法是2重for遍历string的子序列用assign(s.begin() + i, s.begin() + j) 求子序列，但是貌似复杂度太高。。再想想

2.2重for的复杂度到了O(n^3)，重复的判断太多，改为枚举中心对称点：对于N的string 就有2N-1个中心对称点，从左右开始判断即可，O(n^2)

3.继续WA，还好有样例给出，'\n'符的判断也很重要

/* ID: zhangw31PROG: calfflacLANG: C++*/#include <iostream>#include <cctype>#include <string>#include <cstring>#include <fstream>using namespace std;string s;string b;string answer;string tmp;int t[20005];int cr[20005];int len, maxlen, left1, right1;void longPalindrom(int i, int j) {while(i >= 0 && j < len) {if (b[i] != b[j])  break; else {i--; j++;}}if (j - i - 1 > maxlen) {maxlen = j - i - 1; left1 = t[i + 1]; right1 = t[j - 1];}}int countalpha(string s) {int len = s.length();int a = 0;for (int i = 0; i < len; i++) {if (isalpha(s[i])) a++;}return a;}int main() {ifstream fin("calfflac.in");ofstream fout("calfflac.out");while(getline(fin, tmp)) {s += tmp;cr[s.length()-1] = 1;}len = s.length();int counter = 0;for (int i = 0; i < len; i++) {if (isalpha(s[i])) {b += tolower(s[i]);t[counter++] = i;}}len = b.length();for (int i = 0; i < len; ++i) {longPalindrom(i-1, i+1);longPalindrom(i, i+1);}fout << maxlen << endl;for (int i = left1; i <= right1; ++i) {fout << s[i];if (cr[i] == 1) fout << endl;}if(cr[right1] != 1) fout << endl;}

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