HDU 5371 Hotaru's problem Manacher+尺取法

题意：给你一个序列，求最长的两段回文子串，要求他们共用中间的一半。

思路：利用Manacher求出p[i]表示的当前位置的最长回文串长度，然后把每一个长度大于等于2的回文串的左区间和右区间分别放到两个数组里面，由于做manacher时添加了特殊的数字，所以处理的时候稍微注意一下。

　　然后把左右区间按照左端点排序，接着根据尺取法来找答案中的那一段重合的部分。对于每一段右区间R[i]而言，只有L[j].right<=R[i].right && L[j].left<=R[i].right 就是有效的，对于这样的情况所得到的有效值就是L[j].right-R[i].left+1， 当然，当L[j].left>R[i].right时，就不能够再处理R[i]了，而是去处理R[i+1]。

#include <iostream>
#include <cstdio>
#include <algorithm>
#define LL long long
#define MAXN 100005
using namespace std;
int n;
LL s[MAXN * 3];
int p[MAXN];
int f[MAXN];
int m;
struct Node{
int left, right;
int mm;
Node(int left = 0, int right = 0):left(left), right(right){
mm = right - left + 1;
};
bool operator < (const Node & b) const{
return left < b.left;
}
};
vector<Node> L, R;
void manacher(){
int res = 0, id = 0;
for(int i = 1; i <= m; i++) {
if(res > i){
p[i] = min(p[2 * id - i], res - i);
}
else{
p[i] = 1;
}
//p[i] = mx > i? min(mp[2*id-i], mx-i): 1;
while(s[i + p[i]] == s[i - p[i]]){
p[i]++;
}
//while(s[i+mp[i]] == s[i-mp[i]]) mp[i]++;
if(i + p[i] > res) {
res = i + p[i];
id = i;
}
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif // OPEN_FILE
int T;
scanf("%d", &T);
int cas = 1;
while(T--){
scanf("%d", &n);
// char ch;
m = 1;
s[m++] = -2;
s[m++] = -1;
//LL x;
for(int i = 0; i < n; i++){
scanf("%I64d", &s[m++]);
s[m++] = -1;
}
s[m++] = -2;
m--;
manacher();
L.clear();
R.clear();
for(int i = 4; i <= m; i += 2){
if(s[i] != -1 || p[i] == 1) continue;
int x = (i / 2) - 1;
int y = x - ((p[i] - 1) / 2) + 1;
L.push_back(Node(y, x));
x++;
y = x + ((p[i] - 1) / 2) - 1;
R.push_back(Node(x, y));
//printf("%d ", p[i] -1);
}
sort(L.begin(), L.end());
sort(R.begin(), R.end());
/*for(int i = 0; i < R.size(); i++){
printf("%d %d\n", R[i].left, R[i].right);
}*/
int t = 0;
int ans = 0;
for(int i = 0; i < R.size(); i++){
while(t < L.size() && L[t].left <= R[i].left){
if(R[i].left <= L[t].right && L[t].right <= R[i].right){
ans = max(ans, L[t].right - R[i].left + 1);
}
t++;
}
}
printf("Case #%d: %d\n", cas++, ans * 3);
}
}