Section 5

OnlineJudges/poj/2187.cpp

@@ -0,0 +1,53 @@
+#include <cstdio>
+#include <algorithm>
+inline int max(int a, int b) { return a > b ? a : b; }
+const int N = 50005;
+struct point
+{
+    int x, y;
+    point() {}
+    point(int _x, int _y) : x(_x), y(_y) {}
+}a[N], stk[N];
+#define pArg(x) const point &x
+inline point operator+(pArg(l), pArg(r)) { return point(l.x + r.x, l.y + r.y); }
+inline point operator-(pArg(l), pArg(r)) { return point(l.x - r.x, l.y - r.y); }
+inline int cross(pArg(l), pArg(r)) { return l.x * r.y - l.y * r.x; }
+inline int dot(pArg(l), pArg(r)) { return l.x * r.x + l.y * r.y; }
+inline int len2(pArg(p)) { return p.x * p.x + p.y * p.y; }
+inline bool cmp1(pArg(l), pArg(r)) { return l.x < r.x || (l.x == r.x && l.y < r.y); }
+inline bool cmp2(pArg(l), pArg(r))
+{
+    int det = cross(l - a[0], r - a[0]);
+    return det != 0 ? det > 0 : len2(l - a[0]) < len2(r - a[0]);
+}
+#define nxt(x) ((x) == n - 1 ? 0 : (x) + 1)
+#define area(p, u, v) cross(u - p, v - p)
+int main()
+{
+    int n;
+    scanf("%d", &n);
+    for (int i = 0; i < n; i++)
+        scanf("%d%d", &a[i].x, &a[i].y);
+    std::swap(a[0], *std::min_element(a, a + n, cmp1));
+    std::sort(a + 1, a + n, cmp2);
+    int top = 0;
+    for (int i = 0; i < n; i++)
+    {
+        while (top >= 2 && cross(a[i] - stk[top - 1], stk[top - 1] - stk[top - 2]) >= 0) top--;
+        stk[top++] = a[i];
+    }
+    stk[top] = stk[0];
+    int res = 0;
+    if (top == 2)
+        res = len2(stk[1] - stk[0]);
+    else for (int i = 0, j = 2; i < n; i++)
+    {
+        while (nxt(j) != i &&
+               area(stk[j], stk[i], stk[i + 1]) <=
+               area(stk[j + 1], stk[i], stk[i + 1]))
+            j = nxt(j);
+        res = max(res, max(len2(stk[j] - stk[i]), len2(stk[j] - stk[i + 1])));
+    }
+    printf("%d", res);
+    return 0;
+}

省选准备/assets/day5/poj2187.cpp

@@ -0,0 +1,53 @@
+#include <cstdio>
+#include <algorithm>
+inline int max(int a, int b) { return a > b ? a : b; }
+const int N = 50005;
+struct point
+{
+    int x, y;
+    point() {}
+    point(int _x, int _y) : x(_x), y(_y) {}
+}a[N], stk[N];
+#define pArg(x) const point &x
+inline point operator+(pArg(l), pArg(r)) { return point(l.x + r.x, l.y + r.y); }
+inline point operator-(pArg(l), pArg(r)) { return point(l.x - r.x, l.y - r.y); }
+inline int cross(pArg(l), pArg(r)) { return l.x * r.y - l.y * r.x; }
+inline int dot(pArg(l), pArg(r)) { return l.x * r.x + l.y * r.y; }
+inline int len2(pArg(p)) { return p.x * p.x + p.y * p.y; }
+inline bool cmp1(pArg(l), pArg(r)) { return l.x < r.x || (l.x == r.x && l.y < r.y); }
+inline bool cmp2(pArg(l), pArg(r))
+{
+    int det = cross(l - a[0], r - a[0]);
+    return det != 0 ? det > 0 : len2(l - a[0]) < len2(r - a[0]);
+}
+#define nxt(x) ((x) == n - 1 ? 0 : (x) + 1)
+#define area(p, u, v) cross(u - p, v - p)
+int main()
+{
+    int n;
+    scanf("%d", &n);
+    for (int i = 0; i < n; i++)
+        scanf("%d%d", &a[i].x, &a[i].y);
+    std::swap(a[0], *std::min_element(a, a + n, cmp1));
+    std::sort(a + 1, a + n, cmp2);
+    int top = 0;
+    for (int i = 0; i < n; i++)
+    {
+        while (top >= 2 && cross(a[i] - stk[top - 1], stk[top - 1] - stk[top - 2]) >= 0) top--;
+        stk[top++] = a[i];
+    }
+    stk[top] = stk[0];
+    int res = 0;
+    if (top == 2)
+        res = len2(stk[1] - stk[0]);
+    else for (int i = 0, j = 2; i < n; i++)
+    {
+        while (nxt(j) != i &&
+               area(stk[j], stk[i], stk[i + 1]) <=
+               area(stk[j + 1], stk[i], stk[i + 1]))
+            j = nxt(j);
+        res = max(res, max(len2(stk[j] - stk[i]), len2(stk[j] - stk[i + 1])));
+    }
+    printf("%d", res);
+    return 0;
+}

省选准备/省选基础算法.tex

@@ -476,4 +476,9 @@ Manacher 模板题
 \paragraph{练习题}
 \subparagraph{\href{http://poj.org/problem?id=3348}{POJ3348} - Cows} 模板题，注意公式要背对
 \codeinput[Cows]{assets/day5/POJ3348.cpp}
+\subsection{旋转卡壳} 你知道吗？旋转卡壳有16种读法。\\
+	首先有个很直观的结论：最远点对的两点一定在凸包上。我们的思路就是枚举凸包上的所有边，对每一条边找出凸包上离该边最远的顶点，计算这个顶点到该边两个端点的距离，并记录最大的值。当我们逆时针枚举边的时候，最远点的变化也是逆时针的，这样就可以不用从头计算最远点，而可以紧接着上一次的最远点继续计算，于是我们得到了$O(n)$的算法。
+\paragraph{练习题}
+\subparagraph{\href{http://poj.org/problem?id=2187}{POJ2187} - Beauty Contest} 模板题，注意符号要写对
+\codeinput[Beauty Contest]{assets/day5/POJ2187.cpp}
 \end{document}