题目链接

题目大意

平面上给定两个点集，判定两个点集分别形成的凸多边形能否通过旋转、平移重合。

点集大小 $\le$ $10^5$，坐标范围 [0, $10^8$ ].

思路

题意很明显，先求出凸包再判断两凸包是否同构。这里用的 $Andrew$ 算法求。判同构的话，将俩凸包都转化成字符串的形式，用 $kmp$ 去匹配从而判断该串中是否存在另外一个凸包所对应的字符串。

code

#include <bits/stdc++.h>
#define int long long
#define ll long long
#define pii pair<int, int>

using namespace std;
const int N = 1e5 + 100, M = 2e5 + 100;
int n, m, id;
int tmp1, tmp2, minx, ans;
pair<int, int> s[M];
pair<int, int> t[M];
int nxt[M];

struct Point
{
    int x, y;
} tmp, st;
Point a[N];
vector<Point> v;
vector<Point> e[3];

int dop(Point a, Point b)
{
    return a.x * b.x + a.y * b.y;
}
int crp(Point a, Point b)
{
    return a.x * b.y - a.y * b.x;
}
int len(Point a, Point b)
{
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
Point mins(Point a, Point b)
{
    Point c;
    c.x = b.x - a.x;
    c.y = b.y - a.y;
    return c;
}
bool cmp1(Point a, Point b)
{
    if (a.x == b.x)
    {
        return a.y < b.y;
    }
    return a.x < b.x;
}
bool cmp(Point a, Point b)
{
    int sum1 = crp(mins(st, a), mins(st, b));
    if (sum1 == 0.0)
    {
        if (a.x == b.x)
        {
            return a.y < b.y;
        }
        return a.x < a.x;
    }
    return sum1 > 0.0;
}
void get_nxt()
{
    int i = 0, j = -1;
    nxt[0] = -1;
    while (i < 2 * e[1].size())
    {
        if (j == -1 || s[i] == s[j])
        {
            i++;
            j++;
            nxt[i] = j;
        }
        else
        {
            j = nxt[j];
        }
    }
}
bool kmp()
{
    int i = 0, j = 0;
    while (i < 2 * e[1].size())
    {
        if (j == -1 || s[i] == t[j])
        {
            i++;
            j++;
        }
        else if (j == e[2].size())
        {
            return true;
        }
        else
        {
            j = nxt[j];
        }
    }
    return false;
}

void solve()
{
    cin >> n >> m;
    for (int id1 = 1; id1 <= 2; id1++)
    {
        for (int i = 1; i <= n; i++)
        {
            cin >> tmp1 >> tmp2;
            tmp.x = tmp1, tmp.y = tmp2;
            a[i] = tmp;
        }
        minx = 0x3f3f3f3f;
        sort(a + 1, a + n + 1, cmp1);
        st.x = a[1].x, st.y = a[1].y;
        id = 1;
        for (int i = 2; i <= n; i++)
        {
            v.push_back(a[i]);
        }
        sort(v.begin(), v.end(), cmp);
        e[id1].push_back(a[id]);
        for (auto x : v)
        {
            if (e[id1].size() <= 1)
            {
                e[id1].push_back(x);
                continue;
            }
            bool fl = false;
            while (e[id1].size() >= 2)
            {
                int sum1 = crp(mins(e[id1][e[id1].size() - 1], e[id1][e[id1].size() - 2]), mins(e[id1][e[id1].size() - 1], x));
                if (sum1 >= 0.0)
                {
                    e[id1].pop_back();
                }
                else
                {
                    e[id1].push_back(x);
                    fl = true;
                    break;
                }
            }
            if (!fl)
            {
                e[id1].push_back(x);
            }
        }
        n = m;
        v.clear();
    }

    if (e[1].size() != e[2].size())
    {
        cout << "NO\n";
        return;
    }
    for (int i = 0; i < e[1].size(); i++)
    {
        s[i] = {(e[1][i].x - e[1][(i + 1) % e[1].size()].x) * (e[1][i].x - e[1][(i + 1) % e[1].size()].x) + (e[1][i].y - e[1][(i + 1) % e[1].size()].y) * (e[1][i].y - e[1][(i + 1) % e[1].size()].y), dop(mins(e[1][(i + 1) % e[1].size()], e[1][i]), mins(e[1][(i + 1) % e[1].size()], e[1][(i + 2) % e[1].size()]))};
    }
    for (int i = 0; i < e[1].size(); i++)
    {
        s[i + e[1].size()] = s[i];
    }
    for (int i = 0; i < e[2].size(); i++)
    {
        t[i] = {(e[2][i].x - e[2][(i + 1) % e[2].size()].x) * (e[2][i].x - e[2][(i + 1) % e[2].size()].x) + (e[2][i].y - e[2][(i + 1) % e[2].size()].y) * (e[2][i].y - e[2][(i + 1) % e[2].size()].y), dop(mins(e[2][(i + 1) % e[2].size()], e[2][i]), mins(e[2][(i + 1) % e[2].size()], e[2][(i + 2) % e[2].size()]))};
    }
    get_nxt();
    cout << (kmp() ? "YES" : "NO");
}

signed main()
{
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    int tp = 1;
    // cin >> t;
    while (tp--)
        solve();
    return 0;
}