BestCoder Round #49 ($) HDOJ5340 Three Palindromes(暴力)

Problem Description

Can we divided a given string S into three nonempty palindromes?



Input

First line contains a single integer T≤20 which
denotes the number of test cases.

For each test case , there is an single line contains a string S which only consist of lowercase English letters.1≤|s|≤20000



Output

For each case, output the "Yes" or "No" in a single line.



Sample Input

2
abc
abaadada

Sample Output

Yes
No

问一个字符串能不能分成三部分回文

两边枚举字符串 最后判断中间字符串是不是回文就好

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "vector"
using namespace std;
const int maxn = 20010;
char s[maxn];
int len;
vector<int> vec[26];
bool ok(int l, int r)
{
    while(l <= r) {
        if(s[l++] != s[r--]) return false;
    }
    return true;
}
bool judge()
{
    for(int i = 0; i < vec[s[0] - 'a'].size(); ++i) {
        int u = vec[s[0] - 'a'][i];
        if(u + 2 >= len || !ok(0, u)) continue;
        u++;
        for(int j = 0; j < vec[s[u] - 'a'].size(); ++j) {
            int t = vec[s[u] - 'a'][j];
            if(t < u) continue;
            if(t + 1 >= len || s[t + 1] != s[len - 1] || !ok(u, t)) continue;
            if(ok(t + 1, len - 1)) return true;
        }
    }
    return false;
}
int main(int argc, char const *argv[])
{
    int t;
    scanf("%d", &t);
    while(t--) {
        scanf("%s", s);
        len = strlen(s);
        for(int i = 0; i < 26; ++i)
            vec[i].clear();
        for(int i = 0; i < len; ++i)
            vec[s[i] - 'a'].push_back(i);
        if(judge()) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}