题意：
给出一个字符串，问这个字符串能不能分割成三段回文串。
思路：
马拉车判断回文串，然后枚举两边回文串的左右端点，看看中间剩下的部分是不是回文串，是则有解。
代码：
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <map>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <string>
#include <sstream>
#define pb push_back
#define X first
#define Y second
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pii pair<int,int>
#define qclear(a) while(!a.empty())a.pop();
#define lowbit(x) (x&-x)
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define mst(a,b) memset(a,b,sizeof(a))
#define cout3(x,y,z) cout<<x<<" "<<y<<" "<<z<<endl
#define cout2(x,y) cout<<x<<" "<<y<<endl
#define cout1(x) cout<<x<<endl
#define IOS std::ios::sync_with_stdio(false)
#define SRAND srand((unsigned int)(time(0)))
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
using namespace std;
const double PI=acos(-1.0);
const int INF=0x3f3f3f3f;
const ll INFF=0x3f3f3f3f3f3f3f3f;
const ll mod=1e9+9;
const double eps=1e-5;
const int maxn=20005;
const int maxm=20005;
const int base=27;
int T;
char str[maxn];
char ma[maxn<<1];
int mp[maxn<<1];
void manacher(char s[],int len) {
int l=0;
ma[l++]='$';
ma[l++]='#';
for(int i=0; i<len; i++) {
ma[l++]=s[i];
ma[l++]='#';
}
ma[l]=0;
int mx=0,id=0;
for(int i=0; i<l; i++) {
mp[i]= mx>i?min(mp[2*id-i],mx-i):1;
while(ma[i+mp[i]]==ma[i-mp[i]])
mp[i]++;
if(i+mp[i]>mx) {
mx=i+mp[i];
id=i;
}
}
}
vector<int> pre;
vector<int> suf;
void solve() {
sd(T);
while(T--) {
scanf("%s",str);
int len=strlen(str);
manacher(str,len);
pre.clear();
suf.clear();
int lim=2*len+2;
for(int i=2; i<lim-1; i++) {
if(mp[i]==i)pre.pb(i+mp[i]-1);
if(i+mp[i]-1==lim-1)suf.pb(i-(mp[i]-1));
}
bool ok=0;
for(int i=0;i<pre.size();i++){
for(int j=0;j<suf.size();j++){
if(pre[i]>=suf[j])continue;
int mid=(pre[i]+suf[j])>>1;
int r=mid-pre[i];
if(mp[mid]>=r){
ok=1;
break;
}
}
if(ok)break;
}
if(ok)
printf("Yes\n");
else
printf("No\n");
}
return ;
}
int main() {
#ifdef LOCAL
freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
#else
// freopen("","r",stdin);
// freopen("","w",stdout);
#endif
solve();
return 0;
}