QOJ3225 Snake

等价于对于折线每个端点，都能找到一条直线使得所有之前和之后的点分立两侧，在每个点处极角排序 + 双指针即可。

#include <stdio.h>
#include <algorithm>
typedef long long ll;

const int MAXN = 1010;
int n, tot;

struct point{
    ll x, y;
    int id;
};
point a[MAXN], b[MAXN];

point operator - (point p, point q) {
    point res = p;
    res.x -= q.x, res.y -= q.y;
    return res;
}

ll operator * (point p, point q) {
    return p.x * q.y - q.x * p.y;
}

int quad(point p) {
    if (p. x == 0) {
        if (p.y > 0) return 3;
        else return 7;
    } else if (p.x > 0) {
        if (p.y > 0) return 2;
        else if (p.y == 0) return 1;
        else return 8;
    } else {
        if (p.y > 0) return 4;
        else if (p.y == 0) return 5;
        else return 6;        
    }
}

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%lld%lld", &a[i].x, &a[i].y);
        a[i].id = i;
    }
    for (int i = 1; i <= n; ++i) {
        tot = 0;
        for (int j = 1; j <= n; ++j) {
            if (i == j) continue;
            b[tot++] = a[j] - a[i];
        }
        std::sort(b, b + tot, [](point p, point q){return quad(p) == quad(q) ? p * q > 0 : quad(p) < quad(q);});
        int cnt1 = 0, cnt2 = 0;//cnt1 是 id 比 i 小的点的个数，cnt2 是 id 比 i 大的点的个数
        bool flag = false;
        for (int l = 0, r = -1; l < tot; ) {
            while (r + 1 < l + tot && b[l] * b[(r + 1) % tot] >= 0) {
                r++;
                if (b[r % tot].id < i) cnt1++;
                else cnt2++;
            }
            if ((cnt1 == i - 1 && !cnt2) || (cnt2 == n - i && !cnt1)) {
                flag = true;
                break;
            }
            if (b[l].id < i) cnt1--;
            else cnt2--;
            l++;
        }
        if (!flag) {
            printf("Impossible");
            return 0;
        }
    }
    printf("Possible");
    return 0;
}

标签：return,point,int,QOJ3225,else,Snake,cnt1,cnt2
来源： https://www.cnblogs.com/znk161223/p/16584006.html