[POI2007][对称轴osi][计算几何 + KMP]

/*
题目：对称轴osi
题目来源：POI2007
    http://www.zybbs.org/JudgeOnline/problem.php?id=1100
题目内容或思路：
    思路参考自：http://hi.baidu.com/nplusnplusnplu/blog/item/d260baef2e9e9c5879f055cb.html
    給定一個不自交的多邊形判斷對稱軸有幾條。多邊形邊=100000。

        將多邊形的角和邊字符化，假設存儲于字符串S，SS表示兩個S拼接而成的字符串，
    S'表示S翻轉後的串，成那麼原文題等價于：求在字符串SS（去掉末尾字符）中S'出現
    的次數。KMP、Sunday、BM什麽的都可以。
做题日期：2011.8.4
*/
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <algorithm>
using namespace std;

#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define clr(a, b) memset(a, b, sizeof(a))
#define SZ(a) ((int)a.size())
#define PB push_back
#define MP make_pair
#define inf 0x3f3f3f3f
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef long long ll;

#define sqr(x) ((x)*(x))
const int N = 100010;
int tot, n, fail[N * 2];
ll mch[N * 4], pat[N * 2];

struct cpoint { int x, y; }p[N];

void makefail(ll *t, int lt) {
    t--;
    for (int i = 1, j = 0; i <= lt + 1; ++i, ++j) {
        fail[i] = j;
        while (j > 0 && t[i] != t[j]) j = fail[j];
    }
}

int kmp(ll *s, int ls, ll *t, int lt) {
    --s, --t; int cnt = 0;
    for (int i = 1, j = 1; i <= ls; ++i, ++j) {
        while (j > 0 && s[i] != t[j]) j = fail[j];
        if (j == lt) {
            cnt++;
            j = fail[lt + 1] - 1;
        }
    }
    return cnt;
}

ll dissqr(cpoint u, cpoint v) {
    return sqr((ll)u.x - v.x) + sqr((ll)u.y - v.y);
}

ll cross(cpoint p0, cpoint p1, cpoint p2) {
    return (ll)(p1.x - p0.x) * (p2.y - p0.y) -
            (ll)(p2.x - p0.x) * (p1.y - p0.y);
}

void solve() {
    scanf("%d", &n);
    forn (i, n) scanf("%d%d", &p[i].x, &p[i].y);
    tot = 0; p[n] = p[0]; p[n + 1] = p[1];
    forn (i, n) {
        pat[tot++] = dissqr(p[i], p[i + 1]) | 1LL << 62;
        pat[tot++] = cross(p[i + 1], p[i], p[i + 2]);
    }
    memcpy(mch, pat, tot * sizeof(pat[0]));
    memcpy(mch + tot, pat, tot * sizeof(pat[0]));
    reverse(pat, pat + tot);
//    forn (i, tot + tot) printf("%lld ", mch[i]); puts("");
//    forn (i, tot) printf("%lld ", pat[i]); puts("");
    makefail(pat, tot);
    int ans = kmp(mch, tot * 2 - 1, pat, tot);
    printf("%d\n", ans);
}

int main() {
#ifdef CHEN_PC
    freopen("in", "r", stdin);
#endif
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        solve();
    }
    return 0;
 }