2014 Multi-University Training Contest 8 1002

 http://acm.hdu.edu.cn/showproblem.php?pid=4946 

题目给你n个点，可以有重复的点。每个点有速度，问它是否是无限点，具体看题目。。

这个题想法很简单，先找出最大的速度，在这里会有重复的最大速度。先去重，然后把所有最大速度的点求凸包，凸包上的点的位置为1，当然，记得标记之前去重的点，因为去重的点也有可能在凸包上，这些点不能为1。现在很清晰了。

第一次敲凸包，刚开始看不懂凸包的模板，自己太弱，要再努力啊。

Area of Mushroom

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 733    Accepted Submission(s): 138

Problem Description

Teacher Mai has a kingdom with the infinite area.

He has n students guarding the kingdom.

The i-th student stands at the position (x_i,y_i), and his walking speed is v_i.

If a point can be reached by a student, and the time this student walking to this point is strictly less than other students, this point is in the charge of this student.

For every student, Teacher Mai wants to know if the area in the charge of him is infinite.

 

Input

There are multiple test cases, terminated by a line "0".

For each test case, the first line contains one integer n(1<=n<=500).

In following n lines, each line contains three integers x_i,y_i,v_i(0<=|x_i|,|y_i|,v_i<=10^4).

 

Output

For each case, output "Case #k: s", where k is the case number counting from 1, and s is a string consisting of n character. If the area in the charge of the i-th student isn't infinite, the i-th character is "0", else it's "1".

 

Sample Input

30 0 31 1 22 2 10

 

Sample Output

Case #1: 100

 

Source

2014 Multi-University Training Contest 8

 

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <vector>#include <algorithm>#define MAXN 1000using namespace std;struct P{int x,y,v,order;}point[MAXN],tubao[MAXN];int pos[MAXN];bool same(P a,P b){    if(a.x==b.x&&a.y==b.y&&a.v==b.v)        return true;    return false;}bool cmp(P a, P b){    return a.v>b.v;}bool cmp_x(const P& p, const P& q){    if(p.x!=q.x) return p.x<q.x;    return p.y<q.y;}P jian(P a, P b){    P c;    c.x=a.x-b.x;    c.y=a.y-b.y;    return c;}int chaji(P a, P b) { return (a.x*b.y)-(a.y*b.x); }vector<P> convex_hull(P* ps, int n){    sort(ps,ps+n,cmp_x);    int k = 0;    vector<P> qs(n*2);    for(int i=0;i<n;i++){        while(k>1&& chaji ( jian(qs[k-1],qs[k-2]), jian(ps[i],qs[k-1]) ) < 0) k--;        qs[k++]=ps[i];    }    for(int i=n-2,t=k;i>=0;i--){        while(k>t&& chaji ( jian(qs[k-1],qs[k-2]), jian(ps[i],qs[k-1]) ) < 0) k--;        qs[k++]=ps[i];    }    qs.resize(k-1);    return qs;}int main(){    int m=1,n;    while( ~scanf("%d",&n) ) {        memset(pos,0,sizeof(pos));        if(n == 0) break;        for(int i = 0; i < n; i++) {            scanf("%d%d%d",&point[i].x,&point[i].y,&point[i].v);            point[i].order=i;        }        printf("Case #%d: ",m++);        sort(point,point+n,cmp);        int v=point[0].v;        //如果最大的v就是0，则所有都为0        if(v==0){            for(int i=0;i<n;i++)                printf("0");            printf("\n");            continue;        }        for(int i = 0; i < n; i++){            //cout<<point[i].x<<" "<<point[i].y<<" "<<point[i].v<<" "<<point[i].order<<endl;            for(int j = 0; j < n && j != i; j++){                if(same(point[i],point[j])) {                    point[j].v=-1000000;                    pos[point[j].order]=-1;                    pos[point[i].order]=-1;                }            }        }        int k=0;        for(int i = 0; i < n; i++){            if(point[i].v==v) tubao[k++]=point[i];        }        if(k==1){            if(pos[point[0].order]!=-1)                pos[point[0].order]=1;            else pos[point[0].order]=0;            for(int i = 0; i < n; i++)                if(pos[i]==-1) pos[i]=0;            for(int i = 0; i < n; i++)                printf("%d",pos[i]);            printf("\n");            continue;        }        vector<P> p=convex_hull(tubao,k);        for(int i = 0; i < p.size(); i++){            if(pos[p[i].order]!=-1) pos[p[i].order]=1;        }        for(int i=0;i<n;i++)            if(pos[i]==-1) pos[i]=0;        for(int i = 0; i < n; i++)            printf("%d",pos[i]);        printf("\n");    }    return 0;}