[POI2007] OSI-Axes of Symmetry

Description

Luogu3454
BZOJ1100

Solution

把边和角顺次放入一个数组里，如果一个点为中心的回文串的半径大于(n)，那就一定是一个对称轴。跑一遍manacher就行。

Code

#include <algorithm>
#include <cmath>
#include <cstdio>
using std::max;

typedef long double ld;
typedef long long ll;
const int N = 4e5 + 10;
const double eps = 1e-8;

struct point {
  ll x, y;
  point(ll x = 0, ll y = 0) : x(x), y(y) {}
} p[N];
int n, ans[N];
struct state {
  double x;
  ll md;
  state(double x = 0, ll md = 0) : x(x), md(md) {}
  bool operator==(const state &a) const {
    return md == a.md && fabs(x - a.x) < eps;
  }
} f[N];

inline double dis(const point &x, const point &y) {
  return sqrt((x.x - y.x) * (x.x - y.x) + (x.y - y.y) * (x.y - y.y));
}
inline double cross(const point &x, const point &y, const point &z) {
  return 1.0 * (y.x - x.x) * (z.y - y.y) - 1.0 * (z.x - y.x) * (y.y - x.y);
}

int main() {
#ifdef WYXLOCAL
  freopen("tmp.in", "r", stdin);
  freopen("tmp.out", "w", stdout);
#endif  // WYXLOCAL
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      scanf("%lld%lld", &p[i].x, &p[i].y);
    }
    p[0] = p[n];
    p[n + 1] = p[1];
    for (int i = 0; i < n; ++i) {
      f[i * 2] = state(dis(p[i], p[i + 1]), 1);
      f[i * 2 + 1] = state(cross(p[i], p[i + 1], p[i + 2]), 2);
    }
    for (int i = 0; i < n * 2; ++i) f[n * 2 + i] = f[i];
    f[n * 4] = state(0, 0);
    int mxr = 0, pos = 0;
    for (int i = 0; i < 4 * n; ++i) {
      if (i < mxr)
        ans[i] = std::min(ans[2 * pos - i], mxr - i + 1);
      else
        ans[i] = 1;
      while (ans[i] <= i && f[i + ans[i]] == f[i - ans[i]]) ans[i]++;
      if (i + ans[i] - 1 > mxr) {
        mxr = i + ans[i] - 1;
        pos = i;
      }
    }
    int cnt = 0;
    for (int i = 0; i < 4 * n; ++i)
      if (ans[i] > n) cnt++;
    printf("%d
", cnt / 2);
  }
  return 0;
}