【bzoj】1100: [POI2007]对称轴osi

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【算法】

如果多边形关于某个对称轴对称，则从这一点开始的两边的角也是一一对应的（角指的是两条边及夹角），所以用如果将多边形剪成一条线，然后复制一次，在里面找到长度大于等于n的回文串就行了。

计算几何和字符串的结合。

【注意】

关于角的相等可以计算出角度和长度，也可以比较叉积和点积（必须两个）是否相等。

【代码】

写的是Manacher，速度还行，暂列第一页。（忽视代码长度……）

/**************************************************************    Problem: 1100    User: zhengly123    Language: C++    Result: Accepted    Time:1460 ms    Memory:15336 kb****************************************************************/ #include <iostream>#include <stdio.h>#include <stdlib.h>#include <algorithm>#include <math.h>#include <string.h>#define MEM(a) memset(a,0,sizeof(a))using namespace std;typedef long long LL;const int INF = 999999999;const int N = 400040;const double eps = 1e-8; int G(){    int k = 1, tmp = 0;    char ch = getchar();    for (; ch > '9' || ch < '0'&&ch != '-'; ch = getchar());    if (ch == '-') k = -1, ch = getchar();    for (; '0' <= ch && ch <= '9'; ch = getchar()) tmp = tmp * 10 + int(ch) - 48;    return tmp*k;} int zero(double a){    if (fabs(a) < eps) return 0;    else return (a>0) ? 1 : -1;} struct point{    double x, y;    point(){}    point(double a, double b){ x = a; y = b; }    friend point operator+(const point&a, const point&b){ return point(a.x + b.x, a.y + b.y); }    friend point operator -(const point&a, const point&b){ return point(a.x - b.x, a.y - b.y); }    friend point operator *(const double&k, const point&a){ return point(k*a.x, k*a.y); }//数乘    friend point operator /(const point&a, const double&b){ return point(a.x / b, a.y / b); }    friend double operator *(const point&a, const point&b){ return a.x*b.x + a.y*b.y; }//点积    friend double operator /(const point&a, const point&b){ return a.x*b.y - a.y*b.x; }//叉积,用^优先级有问题    friend bool operator <(const point&a, const point&b){ return (a.x - b.x < -eps) || (fabs(a.x - b.x) < eps && (a.y - b.y) < -eps); }    friend bool operator ==(const point&a, const point&b){ return zero(a.x - b.x) && zero(a.y - b.y); }    double dis(){ return (sqrt(x*x + y*y)); }    void input(){ /*scanf("%lf%lf",&x,&y)*/x = G(), y = G(); }    void output(){ printf("%lf %lf\n", x, y); }}p[N]; struct angle{    double poi, cro;    bool operator ==(angle a){ return (zero(poi - a.poi) == 0 && zero(cro - a.cro) == 0); }}s[N];int f[N];void work(){    MEM(f); MEM(s);    int i, j, n, id, maxn, nn, ans = 0;    scanf("%d", &n); nn = 4 * n - 1;    for (i = 1; i <= n; ++i) p[i].input();    p[0] = p[n], p[n + 1] = p[1], s[0].cro = s[0].poi = INF;    for (i = 1; i <= n; ++i)    {        s[2 * i - 1].poi = (p[i - 1] - p[i]) * (p[i + 1] - p[i]);        s[2 * i - 1].cro = (p[i - 1] - p[i]) / (p[i + 1] - p[i]);        s[2 * i - 1 + n * 2] = s[2 * i - 1];    }    for (f[1] = id = maxn = i = 1; i <= nn; ++i)    {        if (maxn > i) f[i] = min(f[2 * id - i], maxn - i);        else f[i] = 1;        for (; s[i + f[i]] == s[i - f[i]]; ++f[i]);        if (f[i] + i>maxn) maxn = f[i] + i, id = i;    }    for (i = n + 1; i <= 3 * n; ++i) if (f[i] >= n - (!(n & 1))) ++ans;    printf("%d\n", ans / 2);} int main(){    int T;    scanf("%d", &T);    while (T--) work();    return 0;}

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