poj 1755 Triathlon（半平面交解可行域）

Triathlon

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 4990 Accepted: 1251

Description

Triathlon is an athletic contest consisting of three consecutive sections that should be completed as fast as possible as a whole. The first section is swimming, the second section is riding bicycle and the third one is running. 

The speed of each contestant in all three sections is known. The judge can choose the length of each section arbitrarily provided that no section has zero length. As a result sometimes she could choose their lengths in such a way that some particular contestant would win the competition. 

Input

The first line of the input file contains integer number N (1 <= N <= 100), denoting the number of contestants. Then N lines follow, each line contains three integers Vi, Ui and Wi (1 <= Vi, Ui, Wi <= 10000), separated by spaces, denoting the speed of ith contestant in each section.

Output

For every contestant write to the output file one line, that contains word "Yes" if the judge could choose the lengths of the sections in such a way that this particular contestant would win (i.e. she is the only one who would come first), or word "No" if this is impossible.

Sample Input

910 2 610 7 35 6 73 2 76 2 63 5 78 4 610 4 21 8 7

Sample Output

YesYesYesNoNoNoYesNoYes

Source

Northeastern Europe 2000

题意：举行一项铁人三项运动，给出每个人在每个项目的速度，问对于每一个人，是否可以设计一定长度的跑道，使某个人获胜

题解：将3个未知数根据约束关系化成2个，详细题解的话推荐以下博客，已经很详细了，就不再写了

             另外要注意的是，这题精度真的很难控制。。。RP不好就会WA了，eps在1e-16比较保险。。。和别人代码差不多，但是还是不断死在精度上面，真不懂为什么别人1e-8能A，我要1e-16才能A，哎。。。

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#include<stdio.h>#include<math.h>#define eps 1e-8#define N 108#define inf 0xffffffstruct point{    double x,y;}p[1508],temp[1508],tp[1508];double a[N],b[N],c[N];double a1,b1,c1;int all;double cross(struct point p1,struct point p2,struct point p3){    return (p2.x-p1.x)*(p3.y-p1.y)-(p2.y-p1.y)*(p3.x-p1.x);}void init(){    int i;    all=4;    p[1].x=0;   p[1].y=0;    p[2].x=0;   p[2].y=inf;    p[3].x=inf; p[3].y=inf;    p[4].x=inf; p[4].y=0;    p[0]=p[4],p[5]=p[1];    for(i=0;i<=all+1;i++) temp[i]=p[i];}void get_line(int x,int y){    a1=(a[y]-a[x])/(a[x]*a[y]);    b1=(b[y]-b[x])/(b[x]*b[y]);    c1=(c[y]-c[x])/(c[x]*c[y]);}void get_intersect(struct point p1,struct point p2){    double a2=p1.y-p2.y;    double b2=p2.x-p1.x;    double c2=p1.x*p2.y-p2.x*p1.y;    double tmd=a1*b2-a2*b1;    temp[1506].x=(b1*c2-b2*c1)/tmd;    temp[1506].y=(a2*c1-a1*c2)/tmd;}void cut(){    int top=0,i;    for(i=1;i<=all;i++)    {        if(a1*temp[i].x+b1*temp[i].y+c1<eps) tp[++top]=temp[i];        else        {            if(a1*temp[i-1].x+b1*temp[i-1].y+c1<-eps)            {                get_intersect(temp[i-1],temp[i]);                tp[++top]=temp[1506];            }            if(a1*temp[i+1].x+b1*temp[i+1].y+c1<-eps)            {                get_intersect(temp[i+1],temp[i]);                tp[++top]=temp[1506];            }        }    }    all=top;    tp[0]=tp[all],tp[all+1]=tp[1];    for(i=0;i<=all+1;i++) temp[i]=tp[i];}int main(){    int n,i,j,flag;    double area;    //freopen("t.txt","r",stdin);    while(scanf("%d",&n)>0)    {        for(i=0;i<n;i++) scanf("%lf%lf%lf",&a[i],&b[i],&c[i]);        for(i=0;i<n;i++)        {            init();            for(flag=1,j=0;j<n&&flag;j++)            {                if(i==j) continue;                get_line(i,j);                if(a1==0&&b1==0&&c1>-eps) flag=0;                cut();            }            for(area=0,j=2;j<all;j++) area+=cross(temp[1],temp[j],temp[j+1]);            if(fabs(area/2.0)<1e-16||!flag) printf("No\n");            else printf("Yes\n");        }    }    return 0;}