题解 P5735 【【深基7.例1】距离函数】
Suuon_Kanderu
2020-01-20 20:48:52
首先,三角形周长为 $AB+BC+AC$.
其次$(x1,x2)$和 $(y1,y2)$的距离为
∣a−b|= $ \sqrt{((x1-x2)^2+(y1-y2)^2} $
然后就可以~~为所欲为~~
CODE:
```
#include<bits/stdc++.h>
using namespace std;
#define N 25
bool a[N][N][N];
int ans=0;
double lon(double x1,double y1,double x2,double y2){
return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
int main(){
double x1,x2,x3,y1,y2,y3;
cin>>x1>>y1>>x2>>y2>>x3>>y3;
printf("%.2lf",lon(x1,y1,x2,y2)+lon(x1,y1,x3,y3)+lon(x2,y2,x3,y3));
return 0;
}
```