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传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=2084
这道题很容易想到就是一个变种的最长回文字串, 不过回文的规则变成了s[i + p[i]] + s[i - p[i]] == 1 可以用hash 来nlogn, 不过最优是用manacher, 然后有一个非常迷的查空位的方法(‘ ‘ ) 可以看代码
#include <iostream> #include <cstring> #include <algorithm> #include <cstdio> using namespace std; const int maxn = 20000100; int n; long long ans = 0; char s[maxn]; int st[maxn]; void read() { scanf("%d", &n); scanf("%s", s + 1); for(int i = 1; i <= n; ++ i) st[i << 1] = (s[i] == ‘0‘ ? 0 : 2); n <<= 1; ++ n; for(int i = 1; i <= n; i += 2) st[i] = 1; } int p[maxn]; void sov() { int mx, id; mx = 0, id = 0; for(int i = 1; i <= n; i += 2) { if(mx > i) p[i] = min(p[2 * id - i], mx - i); else p[i] = 1; for(; i - p[i] > 0 && i + p[i] <= n && st[i - p[i]] + st[i + p[i]] == 2; ++ p[i]); if(i + p[i] > mx) mx = i + p[i], id = i; } for(int i = 1; i <= n; i += 2) ans += (long long)((p[i] - 1) >> 1); printf("%lld\n", ans); } int main() { //freopen("test.in", "r", stdin); read(), sov(); return 0; }
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原文地址:http://www.cnblogs.com/ianaesthetic/p/4251736.html