不慌不忙、不急不躁

题目链接：uva 1373 - How I Wonder What You Are!#include #include #include #include #include #include using namespace std;const double eps = 1e-6;inline int dcmp (double x) { if (fabs(x) < e

题目链接：uva 1342 - That Nice Euler Circuit欧拉定理：边数-点的个数+2=面数#include #include #include #include using namespace std;const double pi = 4 * atan(1);const double eps = 1e-9;inline int dc

题目链接：uva 1301 - Fishnet#include #include #include #include #include using namespace std;const double pi = 4 * atan(1);const double eps = 1e-14;inline int dcmp (double x) { if (fabs(x)

题目链接：uva 10969 - Sweet Dream枚举两圆，获得交点，这样每个圆就被分成若干段弧，判断每段弧时候可见。#include #include #include #include #include using namespace std;const double pi = 4 * atan(1);const double eps = 1e-1

题目链接：uva 10335 - Ray Inside a Polygon恶心题，注意精度和输出等问题，代码中有标识，后面有一些数据。#include #include #include #include #include using namespace std;const double pi = 4 * atan(1);const double eps =

题目链接：uva 11928 - The Busy Dog#include #include #include #include #include using namespace std;const double pi = 4 * atan(1);const double eps = 1e-6;inline int dcmp (double x) { if (fab

题目链接:uva 800 - Crystal Clear不知道士标程的问题还是精度的问题，在UVAlive上过掉了。#include #include #include #include #include using namespace std;const double pi = 4 * atan(1);const double eps = 1e-9;i

题目链接：uva 1100 - Paperweight#include #include #include #include #include #include using namespace std;const double eps = 1e-8;inline int dcmp (double x) { if (fabs(x) < eps) return 0; e

题目链接：uva 11796 - Dog Distance将一只狗位移向量加到另一只上，即变成点到线段的最短距离。#include #include #include #include using namespace std;const double pi = 4 * atan(1);const double eps = 1e-9;inline int d

题目链接：uva 12304 - 2D Geometry 110 in 1!没什么好说的，根据操作直接处理。#include #include #include #include #include using namespace std;const double pi = 4 * atan(1);const double eps = 1e-9;inlin

题目链接：uva 1308 - Viva Confetti枚举一下两圆，处理出所有弧，然后判断每段弧的中点是否可见，可见的话该弧所在的圆也可见，以及该段弧下面的圆也可见。#include #include #include #include #include using namespace std;const double pi = 4 * atan(1);

题目链接：uva 11178 - Morley's Theorem直接按照题意处理。#include #include #include #include using namespace std;const double pi = 4 * atan(1);const double eps = 1e-9;inline int dcmp (double x

题目链接：uva 11646 - Athletics Track#include #include #include #include using namespace std;const double pi = 4.0 * atan(1.0);int main () { int a, b, cas = 1; char c[10]; while (scanf("%d%s

题目链接：uva 1333 - Model Rocket Height#include #include #include #include using namespace std;const double eps = 1e-9;const double pi = 4 * atan(1);inline int dcmp(double x) { if (fabs(x)

题目链接：uva 10794 - The Deadly Olympic Returns!!!点到线段的距离，终点定无穷远。#include #include #include #include #include #include using namespace std;const double eps = 1e-8;inline int dcmp (do

题目链接：uva 11836 - Star War就三种情况，点到线段，点到面，线段到线段。#include #include #include #include #include #include using namespace std;const double eps = 1e-6;inline int dcmp (double x) { if (f

题目链接：uva 535 - Globetrotter#include #include #include #include #include #include using namespace std;//Lab = R * arccos(cos(wa)*cos(wb)*cos(ja-jb) + sin(wa)*sin(wb))const int maxn

题目链接：uva 11275 - 3D Triangles#include #include #include #include #include using namespace std;const double eps = 1e-9;inline int dcmp (double x) { if (fabs(x) < eps) return 0; else retu

题目链接：uva 1469 - Ardenia#include #include #include #include using namespace std;typedef long long ll;struct Point3 { int x, y, z; Point3 (int x = 0, int y = 0, int z = 0): x(x), y(y),

题目链接：uva 11524 - InCircle#include #include #include #include using namespace std;const int maxn = 3;double cal (double a, double b, double c) { return a + b - c;}int main () { int cas

题目链接：uva 11817 - Tunnelling the Earth#include #include #include #include using namespace std;const double R = 6371009;const double pi = 4 * atan(1.0);struct Point { double x, y, z;}p[2]

题目链接：uva 11437 - Triangle Fun#include #include #include #include #include using namespace std;const int maxn = 3;struct point { double x, y; void input() { scanf("%lf%lf", &x, &y);

题目链接：uva 11800 - Determine the Shape#include #include #include #include using namespace std;const int maxn = 4;const double eps = 1e-6;const char T[6][30] = {"Square", "Rectangle", "Rhombu

题目链接：uva 10075 - Airlines在球面上处理出两点的距离，Floyd处理。#include #include #include #include #include #include //#include using namespace std;/* d = r*sqrt(2-2*(cos(lat1)*cos(lat2)*cos(lon1-

题目链接：uva 503 - Parallelepiped walk恶心题，将三维转成两维，直线距离最短，WA了一天。假设起点在地面，除了考虑经过0，1个面的可能，还要考虑经过两个面到达的可能。后面提供一个生成数据的代码。#include #include #include using namespace std;typedef long long ll;con

题目链接：uva 303 - Pipe枚举射线，然后逐一判断。#include #include #include #include #include using namespace std;const double pi = 4 * atan(1);const double eps = 1e-6;inline int dcmp (double x) {

题目链接：hihoCoder 1231 Border Length解题思路处理处所有交点，按照交点和存在的点分段，如果是圆，则每一段均为弧，如果是多边形的边，则每一段为线段。每段取中间点，判断是否在多边形或者圆中。代码#include <cstdio>#include <cstring>#include <cmath>#include <vector>#include <complex>#

题目链接：hdu 3264 Open-air shopping malls解题思路圆与圆面积交代码#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const double pi = 4 * atan(1.0);const double eps = 1e-8;

题目链接：hdu 4606 Occupy Cities解题思路首先预处理出两两点之间的最短距离，然后二分背包容量，用KM判断是否可行。预处理部分，将所有线段的端点加入考虑，枚举两点之间直线，如果与线段相交则不可以移动。然后用floyd处理出点点之间的最短距离。判断部分，因为有P个士兵，所以对于一个距离，可以根据占领顺序处理出有向边，然后用KM处理至少需要多少个士兵。代码#include <cstdi

题目链接：hdu 4617 Weapon解题思路异面直线之间距离。代码#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <vector>#include <algorithm>using namespace std;const double eps = 1e-8;inline int

题目链接：poj 2546 Circular Area代码#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const double pi = 4 * atan(1.0);const double eps = 1e-8;inline int dcmp (dou

题目连接：hdu 4637 Rain on your Fat brother解题思路题目等价于求线段与雨水的交，t为mazi停下来的时间，妹子移动相对于雨水的移动路线为(x,0)−(x−V1∗t,V∗t)(x,0)-(x-V1*t, V*t)，然后求出线段与每个雨水相交点，计算出进出时间，注意于圆相交的部分只能计算下半圆。 接着做一下区间覆盖，因为雨滴有重叠。代码#include <cstdio>

题目链接：hdu 4793 Collision代码#include <cstdio>#include <cstring>#include <cmath>#include <vector>#include <algorithm>using namespace std;const double pi = 4 * atan(1);const double eps = 1e-6;inline i

题目链接：hdu 4798 Skycity代码#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const double pi = 4 * atan(1.0);int F;double R, L, H, S;double solve (double r, do

题目链接：hdu 5563 Clarke and five-pointed star代码#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const double eps = 1e-4;struct Point { double x, y; Po

题目链接：hdu 4643 GSM代码#include <cstdio>#include <cstring>#include <cmath>#include <vector>#include <algorithm>using namespace std;const double eps = 1e-10;inline int dcmp(double x) { if (fabs(x)<eps)

题目链接：hdu 5473 There was a kingdom解题思路选取的点一定在凸包上，所以对点集做凸包，如果凸包的点个数小于等于K，面积可以取到最大值。否则，枚举起点，做动态规划。dp[i][j]表示到第i个点选取了j个点的最优解。这样的复杂度为o(n2k)o(n^2k)，算上枚举起点总得复杂度为o(n3k)o(n^3k)。但是，我们选取的是k个点，如果有枚举到一次的起点在这k个点上，即可

题目链接：hdu 5476 Explore Track of Poin#include #include #include #include #include #include using namespace std;typedef pair pii;const double pi = 4 * atan(1);const double eps = 1e-6;in

题目链接：uva 10556 - Biometrics旋转角度可长度比例相同即可，注意精度。#include #include #include using namespace std;typedef long long ll;struct Point { ll x, y; void read() { scanf("%lld%lld", &x, &y);

题目链接：uva 313 - Intervals点和圆的切线，线和线的交点。#include #include #include #include #include using namespace std;const double pi = 4 * atan(1);const double eps = 1e-6;inline int dcmp (doub