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  • 【HDOJ】4513 吉哥系列故事——完美队形II

    这题目上学期就看了,不过最近发现可以用马拉车来解,而且还是基本算法。 稍微对回文串成立条件变形一下即可。

      1 /* 4513 */
      2 #include <iostream>
      3 #include <string>
      4 #include <map>
      5 #include <queue>
      6 #include <set>
      7 #include <stack>
      8 #include <vector>
      9 #include <deque>
     10 #include <algorithm>
     11 #include <cstdio>
     12 #include <cmath>
     13 #include <ctime>
     14 #include <cstring>
     15 #include <climits>
     16 #include <cctype>
     17 #include <cassert>
     18 #include <functional>
     19 #include <iterator>
     20 #include <iomanip>
     21 using namespace std;
     22 //#pragma comment(linker,"/STACK:102400000,1024000")
     23 
     24 #define sti                set<int>
     25 #define stpii            set<pair<int, int> >
     26 #define mpii            map<int,int>
     27 #define vi                vector<int>
     28 #define pii                pair<int,int>
     29 #define vpii            vector<pair<int,int> >
     30 #define rep(i, a, n)     for (int i=a;i<n;++i)
     31 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
     32 #define clr                clear
     33 #define pb                 push_back
     34 #define mp                 make_pair
     35 #define fir                first
     36 #define sec                second
     37 #define all(x)             (x).begin(),(x).end()
     38 #define SZ(x)             ((int)(x).size())
     39 #define lson            l, mid, rt<<1
     40 #define rson            mid+1, r, rt<<1|1
     41 
     42 const int maxn = 1e5+5;
     43 int a[maxn], b[maxn*2];
     44 int Len[maxn*2];
     45 
     46 void init(int n) {
     47     int j = 0;
     48     
     49     b[j++] = -1;
     50     b[j++] = 0;
     51     rep(i, 0, n) {
     52         b[j++] = a[i];
     53         b[j++] = 0;
     54     }
     55 }
     56 
     57 void manacher(int m) {
     58     int n = 2*m+1;
     59     int p = 0, p0 = 0;
     60     int lid, rid;
     61     
     62     rep(i, 1, n+1) {
     63         if (p > i)
     64             Len[i] = min(p-i, Len[2*p0-i]);
     65         else
     66             Len[i] = 1;
     67         
     68         while (b[i+Len[i]] == b[i-Len[i]]) {
     69             if (b[i+Len[i]] > 0) {
     70                 lid = i - Len[i];
     71                 rid = i + Len[i];
     72                 if (b[lid]>b[lid+2] || b[rid]>b[rid-2])
     73                     break;
     74             }
     75             ++Len[i];
     76         }
     77         
     78         if (i+Len[i] > p) {
     79             p = i + Len[i];
     80             p0 = i;
     81         }
     82     }
     83 }
     84 
     85 void solve(int n) {
     86     init(n);
     87     manacher(n);
     88     
     89     int ans = 0;
     90     
     91     n = n*2+1;
     92     rep(i, 1, n) {
     93         ans = max(ans, Len[i]);
     94     }
     95     printf("%d
    ", ans-1);
     96 }
     97 
     98 int main() {
     99     ios::sync_with_stdio(false);
    100     #ifndef ONLINE_JUDGE
    101         freopen("data.in", "r", stdin);
    102         freopen("data.out", "w", stdout);
    103     #endif
    104     
    105     int t;
    106     int n;
    107     
    108     scanf("%d", &t);
    109     while (t--) {
    110         scanf("%d", &n);
    111         rep(i, 0, n)
    112             scanf("%d", &a[i]);
    113         solve(n);
    114     }
    115     
    116     #ifndef ONLINE_JUDGE
    117         printf("time = %d.
    ", (int)clock());
    118     #endif
    119     
    120     return 0;
    121 }
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  • 原文地址:https://www.cnblogs.com/bombe1013/p/4852364.html
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