UVALive

题目链接

分析：
题目要求通过1和2√长度的边把给定的点全部围起来，于是很容易想到求一个凸包，然后长度1个数为max(abs(x1−x2),abs(y1−y2))−min(abs(x1−x2),abs(y1−y2)), 长度为2√的为min(abs(x1−x2),abs(y1−y2))。
需要注意的是叉积会爆int

代码：

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cstdlib>#include <vector>#include <queue>#include <deque>#include <stack>#include <set>#include <cmath>using namespace std;struct Point {    int x, y;    Point (int x = 0, int y = 0) : x(x), y(y) {}    bool operator <(const Point &rhs) const {        if (rhs.x == x) return y < rhs.y;        return x < rhs.x;    }};typedef Point Vector;Vector operator +(Vector A, Vector B) {    return Vector(A.x + B.x, A.y + B.y);}Vector operator -(Point A, Point B) {    return Vector(A.x - B.x, A.y - B.y);}long long Cross(Vector A, Vector B) {    return 1ll * A.x * B.y - 1ll * A.y * B.x;}int ConvexHull(Point *p, int n, Point *ch) {    sort(p, p + n);    int m = 0;    for (int i = 0; i < n; i ++) {        while (m > 1 && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0) m --;        ch[m ++] = p[i];    }    int k = m;    for (int i = n - 2; i >= 0; i --) {        while (m > k && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0) m --;        ch[m ++] = p[i];    }    if (n > 1) m --;    return m;}const int maxn = 1e4 + 5;Point node[maxn], hull[maxn];long long resa, resb;void work(const Point &a, const Point &b) {    int minv = min(abs(a.x - b.x), abs(a.y - b.y));    int maxv = max(abs(a.x - b.x), abs(a.y - b.y));    resb += 1ll * minv;    resa += 1ll * (maxv - minv);}int main() {    int N;    while (~scanf("%d", &N) && N) {        resa = 0, resb = 0;        for (int i = 0; i < N; i++)            scanf("%d%d", &node[i].x, &node[i].y);        int len = ConvexHull(node, N, hull);        for (int i = 1; i < len; i ++) {//            cout<<hull[i - 1].x<<" "<<hull[i - 1].y<<" "<<hull[i].x<<" "<<hull[i].y<<endl;            work(hull[i - 1], hull[i]);        }        work(hull[0], hull[len - 1]);        printf("%lld %lld\n", resa, resb);    }    return 0;}