题目本质：忽略串的变化，只记载26个字母的相关变化。
解决方法：
在上一次与本次的转移过程中，情况并不多，主要取决于本次串的首尾字母，若不是本次的首尾字母，会被置1；如果是的话，分情况接一下并更新。另外应该不会忘记的就是要拿本次串的最长串再更新一下。
复杂度最差时不是O（n2）吗？数据骗了。
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <cmath> 6 #include <ctime> 7 #include <cctype> 8 #include <climits> 9 #include <iostream> 10 #include <iomanip> 11 #include <algorithm> 12 #include <string> 13 #include <sstream> 14 #include <stack> 15 #include <queue> 16 #include <set> 17 #include <map> 18 #include <vector> 19 #include <list> 20 #include <fstream> 21 #define ri readint() 22 #define gc getchar() 23 #define R(x) scanf("%d", &x) 24 #define W(x) printf("%d\n", x) 25 #define init(a, b) memset(a, b, sizeof(a)) 26 #define rep(i, a, b) for (int i = a; i <= b; i++) 27 #define irep(i, a, b) for (int i = a; i >= b; i--) 28 #define ls p << 1 29 #define rs p << 1 | 1 30 using namespace std; 31 32 typedef double db; 33 typedef long long ll; 34 typedef unsigned long long ull; 35 typedef pair<int, int> P; 36 const int inf = 0x3f3f3f3f; 37 const ll INF = 1e18; 38 39 inline int readint() { 40 int x = 0, s = 1, c = gc; 41 while (c <= 32) c = gc; 42 if (c == '-') s = -1, c = gc; 43 for (; isdigit(c); c = gc) 44 x = x * 10 + c - 48; 45 return x * s; 46 } 47 48 const int maxn = 1e5 + 5; 49 50 vector<ll> getLength(string s) { 51 vector<ll> v(26, 0); 52 int len = 1, last = -1; 53 54 rep(i, 0, s.length() - 1) { 55 int now = s[i] - 'a'; 56 if (now == last) { 57 len++; 58 } else { 59 last = now; 60 len = 1; 61 } 62 v[now] = max(v[now], (ll)len); 63 } 64 65 return v; 66 } 67 68 int main() { 69 ios_base::sync_with_stdio(false); 70 cin.tie(0); 71 72 int n; 73 cin >> n; 74 string s; 75 cin >> s; 76 vector<ll> Len = getLength(s); 77 78 rep(i, 1, n - 1) { 79 cin >> s; 80 int prec = s[0] - 'a', sufc = s.back() - 'a'; 81 int prel = 1, sufl = 1; 82 rep(j, 1, s.length() - 1) { 83 if (s[j] - 'a' == prec) prel++; 84 else break; 85 } 86 irep(j, s.length() - 2, 0) { 87 if (s[j] - 'a' == sufc) sufl++; 88 else break; 89 } 90 91 vector<ll> cur = getLength(s); 92 if (prel == s.length()) { 93 rep(j, 0, 25) { 94 if (j == prec) { 95 Len[j] += (Len[j] + 1) * prel; 96 } else { 97 Len[j] = min(Len[j], 1ll); 98 } 99 } 100 } else { 101 rep(j, 0, 25) { 102 if (Len[j]) 103 Len[j] = (prec == j) * prel + (sufc == j) * sufl + 1; 104 Len[j] = max(Len[j], cur[j]); 105 } 106 } 107 108 rep(j, 0, 25) Len[j] = min(Len[j], (ll)1e9); 109 } 110 111 ll ans = -1; 112 rep(i, 0, 25) ans = max(ans, Len[i]); 113 cout << ans << endl; 114 return 0; 115 }
