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如果一个人能统治无穷远处，那么他的速度一定是最大的。除了那几种很坑的数据，比如同一个点速度相同的两个人。永远都是不可能。所以你要处理出来所有速度最大的点，然后用他们构成一个凸包，你把端点的点求出来了，还得判断一下在边上的情况。然后顶点和在边上的点所构成的就是可以到达无穷远处的人。

PS：抄了芳姐的模版。。哈哈哈

Area of Mushroom

3
0 0 3
1 1 2
2 2 1
0

Case #1: 100

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define clearall(A, X) memset(A, X, sizeof(A))

using namespace std;

const int maxn = 2010;

struct point
{
    int x,y;
    int v,id;
    point(int x = 0,int y = 0):x(x),y(y) {}
} p[maxn],ch[maxn],q[maxn],pq[maxn];
int f[maxn];
typedef point pointt;
point operator -(point a,point b)
{
    return point(a.x-b.x,a.y-b.y);
}
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    return x<0?-1:1;
}
int cross(point a,point b)
{
    return a.x*b.y-a.y*b.x;
}
int mul(point a,point b,point c)
{
    return cross(b-a,c-a);
}
double dis(point a)
{
    return sqrt(a.x*a.x+a.y*a.y);
}
bool cmp(point a,point b)
{
    if(mul(p[0],a,b)==0)
        return dis(a-p[0])<dis(b-p[0]);
    return mul(p[0],a,b)>0;
}
int graham(int n)///返回点的个数
{
    int i,k =0 ,top;
    if(n<2) return 0;
    point tmp ;
    for(i = 0; i < n ; i++)
    {
        if(dcmp(p[i].y-p[k].y)<0||(dcmp(p[i].y-p[k].y)==0&&dcmp(p[i].x-p[k].x)<0))
            k = i;
    }
    swap(p[k],p[0]);
    sort(p+1,p+n,cmp);
    ch[0] = p[0];
    ch[1] = p[1];
    top = 1;
    for(i = 2; i < n ; i++)
    {
        while(top>0&&dcmp(mul(ch[top-1],ch[top],p[i]))<=0) top--;
        top++;
        ch[top] = p[i];
    }
    return top;
}
bool cmpp(point a,point b)
{
    if(a.x==b.x) return a.y<b.y;
    return a.x<b.x;
}

int main()
{
    int Case = 1;
    int n;
    while(cin >>n)
    {
        if(!n) break;
        memset(f, 0, sizeof(f));
        int Max = -INF;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d %d %d",&q[i].x, &q[i].y, &q[i].v);
            q[i].id = i;
            if(!q[i].v) f[i] = -2;
            Max = max(Max, q[i].v);
        }
        for(int i = 1; i <= n; i++)
        {
            if(q[i].v != Max) continue;
            for(int j = i+1; j <= n; j++)
            {
                if(q[j].x == q[i].x && q[j].y == q[i].y && q[j].v == q[i].v)
                {
                    f[q[j].id] = -1;
                    f[q[i].id] = -1;
                }
            }
        }
        int t = 0;
        for(int i = 1; i <= n; i++)
            if(q[i].v == Max && f[q[i].id] != -2) p[t++] = q[i];
        int m = graham(t);
        cout<<"Case #"<<Case++<<": ";
        if(m < 2)
        {
            for(int i = 1; i <= n; i++)
            {
                if(f[i] != -1 && f[i] != -2 && q[i].v == Max) cout<<1;
                else cout<<0;
            }
            cout<<endl;
            continue;
        }
        ch[m+1] = ch[0];
        for(int i = 0; i <= m; i++) if(!f[ch[i].id]) f[ch[i].id] = 1;
        for(int i = 1; i <= n; i++)
        {
            if(f[i] || q[i].v != Max) continue;
            for(int j = 0; j <= m; j++)
                if(mul(ch[j], ch[j+1], q[i]) == 0) f[i] = 1;
        }
        for(int i = 1; i <= n; i++)
        {
            if(f[i] == 1) cout<<1;
            else cout<<0;
        }
        cout<<endl;
    }
    return 0;
}

HDU 4946 Area of Mushroom(凸包),布布扣,bubuko.com

HDU 4946 Area of Mushroom(凸包)

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原文地址：http://blog.csdn.net/xu12110501127/article/details/38563603