C++博客-From A Start,As An Acmer-随笔分类-Ural Solution Report

Ural 1588 Jamaica［平面图最短路？］

BUPT-[aswmtjdsj] @ Penalty — Sat, 20 Aug 2011 17:15:00 GMT

http://acm.timus.ru/problem.aspx?space=1&num=1588

题目一点也不不像所说的那么神。。。虽然标准的O(N^2 * logN)的算法不会吧。不过O(N^3)的算法还是能过的。

给你平面上N个点(<=300)。两两连边，求生成网络的路径总长，注意在同一直线上的边长不要重复计算。

O(N^2)预处理出每两点间最短路。然后利用类似floyd的想法N^3枚举出需要处理掉的边，即诸如d[i][j] = d[i][k] + d[k][j]的边，统计所有边长的时候不要加入此种d[i][j]的值。
关键点1：用double的坏处就在这了，自带常矬卡精度。随随便便就可以构造出那种三点构成的三角形最大内角近似180度的情况用来卡这种处理方式。
即三点近似共线的情况用于卡这个式子fabs(d[i][j] - d[i][k] - d[k][j]) < eps。。一开始eps=1e-6果断被卡了，处理到1e-9才在题目给出的数据范围(10^4)下可以不被精度所影响。
“精度”啊“精度”！！！
关键点2：平常不注意cout输出流的格式自适应，用的太习惯以致于察觉不到这种最常见的小事。这你妹。我真是个蒟蒻啊。
cout输出流对于大double数如果不iomanip的话是会自动用科学计数法输出的！！！平常用cout输出二分中间结果的时候经常会看到，现在怎么想不到了呢！！！我真是太弱了。。。
所以对于哪怕题目要求最后输出四舍五入后的整数，也不要“cout<
少年啊，要记住，非常习惯的事物的小小细节也许是你不曾注意到的坑你的地方。世界上的事情不要想当然。无论在什么方面。

BUPT-[aswmtjdsj] @ Penalty 2011-08-21 01:15 发表评论

URAL 1215 Exactness of Projectile Hit 【点到线段的距离和点在多边形内与否的判断】

BUPT-[aswmtjdsj] @ Penalty — Fri, 19 Aug 2011 13:30:00 GMT

题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1215

坑爹的题，让我交了17次才过。。。。这。。
几个关键点。（判断点在“凸”多边形内，果断用叉积判断而非“角和”判断吧。我不知道ural怎么做到的，我把eps开到了1都能把我的“角和”判断卡掉，太nb了。）
1.叉积判断点在多边形内或上
如果旋转叉积过程中出现“正”以及“负”号情况发生，则点在“凸”多边形外。前提，点按顺/逆时针给出。
2.点到线段的距离
根据点积判断点与线段两端点组成的三角形的两个底角（相对于用于判断的点而言）是否均非钝角。
若有一个钝角，则点到线段的距离为点到两端点距离中的小者；若无，则为垂足距离。

（自己写的几何模板不要偷懒。。能用向量绝不解析。。。不然卡精度卡死你。。还有，有标准做法的，千万不要用偷懒做法水过去，不然以后会后悔的）
附代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#define sqr(x) ((x) * (x))
using namespace std;
#define maxn 105
const double eps = 1e-6;
const double pi = acos(-1.0);
int comp(double x)
{
    if(fabs(x) < eps)
        return 0;
    else if(x < -eps)
        return -1;
    else
        return 1;
}
struct point
{
    int x,y;
    point(){}
    point(int a,int b):x(a),y(b){}
    point operator -(const point &p)
    {
        return point(x - p.x,y - p.y);
    }
    int operator *(const point &p)
    {
        return x * p.x + y * p.y;
    }
    int operator ^(const point &p)
    {
        return x * p.y - y * p.x;
    }
    int norm2()
    {
        return sqr(x) + sqr(y);
    }
}p[maxn],o;
int multi(point a,point b,point o)
{
    return (a - o) ^ (b - o);
}
int scalar(point a,point b,point o)
{
    return (a - o) * (b - o);
}
int n;
int main()
{
    int a,b;
    scanf("%d %d",&a,&b);
    o = point(a,b);
    scanf("%d",&n);
    for(int i = 0;i < n;i++)
    {
        scanf("%d %d",&a,&b);
        p[i] = point(a,b);
    }
    bool pos = false,neg = false;
    for(int i = 0;i < n;i++)
    {
        int a = i,b = (i + 1) % n;
        int flag = multi(p[a],p[b],o);
        if(flag > 0 && !pos)
            pos = true;
        if(flag < 0 && !neg)
            neg = true;
    }
    if(!(neg && pos))
    {
        printf("%.3lf\n",0.0);
        return 0;
    }
    double r = 1e10;
    for(int i = 0;i < n;i++)
    {
        int a = i,b = (i + 1) % n;
        double lab = sqrt((double)(p[a] - p[b]).norm2()),lbo = sqrt((double)(o - p[b]).norm2()),lao = sqrt((double)(o - p[a]).norm2());
        int sbao = scalar(p[b],o,p[a]),sabo = scalar(p[a],o,p[b]);
        double bao = acos((double) sbao / (lab * lao)),abo = acos((double)sabo / ( lab * lbo));
        if(comp(bao - pi / 2.0) <= 0 && comp(abo - pi / 2.0) <= 0)
            r = min(r,fabs((double)multi(p[a],p[b],o)) / lab);
        else
            r = min(r,min(lbo,lao));
    }
    printf("%.3lf\n",r * 2.0);
}

BUPT-[aswmtjdsj] @ Penalty 2011-08-19 21:30 发表评论

Ural 1640 Circle of Winter【坑爹题。。。】

BUPT-[aswmtjdsj] @ Penalty — Sun, 14 Aug 2011 13:32:00 GMT

1640. Circle of Winter

Time Limit: 1.0 second
Memory Limit: 64 MB

Lich Sandro declared war to the King of Hell, and now hordes of demons are fighting fiercely against Sandro's army of the undead. Sandro has mastered the Magic of Fire, but the creatures of Hell can't be damaged by fire. That's why Sandro decided to use a “Circle of Winter” spell against them. This spell creates an indestructable circular-shaped ice wall. The demons who are touched by the circumference die immediately; the demons enclosed in a circle stay alive, but can't participate in the fight anymore.

Sandro can teleport to any point of the battlefield instantly and cast a “Circle of Winter” there. An ice wall centered in that point will appear in that case. Note that Sandro's magic skills allow him to cast a wall with a radius not exceeding 10000 metres. Now Sandro wants to choose a point to teleport to and a radius of a spell in such way that at least one demon would be killed and all the rest would be enclosed in a circle.

Input

The first line contains an integer n — the number of demons (1 ≤ n ≤ 100). Demons can be considered as points on the battlefield. The next n lines contain n pairs of space-separated integers (x, y) — the coordinates of the demons relative to the point Sandro occupies before teleportation. Coordinates are given in metres and don't exceed 1000 in absolute value. Each point of the battlefield contains no more than one demon; also a demon can't occupy a point the Sandro occupies before teleportation.

Output

Output 3 real numbers precise up to 10⁻⁹ — the coordinates of a point Sandro should teleport to and the radius of a “Circle of Winter” he should create. Sandro can't teleport to a point that is occupied by a demon. It is guaranteed that the solution always exists.

Samples

input	output
7 1 1 1 5 3 6 5 3 8 0 9 5 5 9	5 4 5
2 0 2 2 0	1 1 1.41421356237309

Problem Author: Dmitry Ivankov (prepared by Alexander Ipatov)
Problem Source: USU Junior Contest, October 2008

Tags: geometry

去年看到的时候就不会。今年会了最小覆盖圆的模板，试着做，结果依然WA，估计是找到的最小覆盖圆和给定的点中的某一个重复了。
仔细一想，试了试，竟然是坑爹题。
题目给你N个二维整点（坐标绝对值不大于1000），让你找一个点使这N个点中至少有一个在圆上，其他均在圆内，且圆半径小于10000，且圆心不与已知点重合。输出圆心坐标和半径（实数，精确到1e-9）

这题坑爹之处在于，不要求输出半径最小解。
那么即可随意构造解了。
只要任意实数解都可，因为题目给出的点都是整点。
这样你就令c.x = 0.1,c.y = 0.1，然后求出这个圆心到这N个点的最远距离，即为半径，然后AC。
我了个去。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
#define maxn 105
const double eps = 1e-9;
double x[maxn],y[maxn],cx,cy,r;
int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 0;i < n;i++)
        scanf("%lf %lf",&x[i],&y[i]);
    cx = cy = 0.1;
    r = 0.0;
    for(int i = 0;i < n;i++)
        r = max(r,sqrt((cx - x[i]) * (cx - x[i]) + (cy - y[i]) * (cy - y[i])));
    printf("%.10lf %.10lf %.10lf\n",cx,cy,r);
}

BUPT-[aswmtjdsj] @ Penalty 2011-08-14 21:32 发表评论