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]。
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #define LL long long 5 #define MAXN 100005 6 using namespace std; 7 int n; 8 LL s[MAXN * 3]; 9 int p[MAXN]; 10 int f[MAXN]; 11 int m; 12 struct Node{ 13 int left, right; 14 int mm; 15 Node(int left = 0, int right = 0):left(left), right(right){ 16 mm = right - left + 1; 17 }; 18 bool operator < (const Node & b) const{ 19 return left < b.left; 20 } 21 }; 22 vector<Node> L, R; 23 void manacher(){ 24 int res = 0, id = 0; 25 for(int i = 1; i <= m; i++) { 26 if(res > i){ 27 p[i] = min(p[2 * id - i], res - i); 28 } 29 else{ 30 p[i] = 1; 31 } 32 //p[i] = mx > i? min(mp[2*id-i], mx-i): 1; 33 while(s[i + p[i]] == s[i - p[i]]){ 34 p[i]++; 35 } 36 //while(s[i+mp[i]] == s[i-mp[i]]) mp[i]++; 37 if(i + p[i] > res) { 38 res = i + p[i]; 39 id = i; 40 } 41 } 42 } 43 int main() 44 { 45 #ifndef ONLINE_JUDGE 46 freopen("in.txt", "r", stdin); 47 //freopen("out.txt", "w", stdout); 48 #endif // OPEN_FILE 49 int T; 50 scanf("%d", &T); 51 int cas = 1; 52 while(T--){ 53 scanf("%d", &n); 54 // char ch; 55 m = 1; 56 s[m++] = -2; 57 s[m++] = -1; 58 //LL x; 59 for(int i = 0; i < n; i++){ 60 scanf("%I64d", &s[m++]); 61 s[m++] = -1; 62 } 63 s[m++] = -2; 64 m--; 65 manacher(); 66 L.clear(); 67 R.clear(); 68 for(int i = 4; i <= m; i += 2){ 69 if(s[i] != -1 || p[i] == 1) continue; 70 int x = (i / 2) - 1; 71 int y = x - ((p[i] - 1) / 2) + 1; 72 L.push_back(Node(y, x)); 73 x++; 74 y = x + ((p[i] - 1) / 2) - 1; 75 R.push_back(Node(x, y)); 76 //printf("%d ", p[i] -1); 77 } 78 sort(L.begin(), L.end()); 79 sort(R.begin(), R.end()); 80 /*for(int i = 0; i < R.size(); i++){ 81 printf("%d %d\n", R[i].left, R[i].right); 82 }*/ 83 int t = 0; 84 int ans = 0; 85 for(int i = 0; i < R.size(); i++){ 86 while(t < L.size() && L[t].left <= R[i].left){ 87 if(R[i].left <= L[t].right && L[t].right <= R[i].right){ 88 ans = max(ans, L[t].right - R[i].left + 1); 89 } 90 t++; 91 } 92 } 93 printf("Case #%d: %d\n", cas++, ans * 3); 94 } 95 }
