85. Three Points On A Line

時間限制1000 ms????內存限制 65536 KB????

題目描述

Given points on a 2D plane, judge whether there're three points that locate on the same line.

輸入格式

The number of test cases T(1≤T≤10)? appears in the first line of input.

Each test case begins with the number of points N(1≤N≤100)?. The following N? lines describe the coordinates (x?i?,y?i?)? of each point, in accuracy of at most 3 decimals. Coordinates are ranged in [?10?4?,10?4?]?.

輸出格式

For each test case, output Yes if there're three points located on the same line, otherwise outputNo.

輸入樣例

2 3 0.0 0.0 1.0 1.0 2.0 2.0 3 0.001 -2.000 3.333 4.444 1.010 2.528

輸出樣例

Yes No

?

三點共線，枚舉所有的3點，求斜率，一樣則存在，做好標記然后退出枚舉即可。還有一種做法是求出一組后，把計算結果存起來，減少計算量。

?

#include<stdio.h> int main(){int t,n,i,j,k;double x[105],y[105];for(scanf("%d",&t);t--;){scanf("%d",&n);for(i=0;i<n;i++)scanf("%lf%lf",x+i,y+i);int sig=0;for(i=0;i<n&&!sig;i++)for(j=i+1;j<n&&!sig;j++){for(k=j+1;k<n&&!sig;k++){if((y[k]-y[i])/(x[k]-x[i])==(y[j]-y[i])/(x[j]-x[i]))sig=1;}}if(sig)printf("Yes\n");else printf("No\n");}return 0; }

?

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