題目大意:如題順時針輸出凸包頂點,並且輸出凸包周長;
注意:(1)題目告訴輸出時,凸包起點任意,但必須輸出兩次;
(2)每兩組數據之間換行,其他就沒什麼了。
解題策略:同上。
/*
UVA 218 Moth Eradication
AC by J_Dark
ON 2013/5/6 16:22
Time 0.042s
*/
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <stack>
using namespace std;
/////////////////////////////////////
struct point{
double x, y;
point(double a, double b){
x = a;
y = b;
}
double Distance(point t){
return sqrt((x-t.x)*(x-t.x) + (y-t.y)*(y-t.y));
}
};
vector<point> p;
vector<int> CH; //存放凸包頂點序號 模擬棧
int testCase, top, cc=0, nodeNum;
/////////////////////////////////////
void Input(){
p.clear();
CH.clear();
CH.resize(nodeNum+5);
double xx, yy;
for(int i=0; i<nodeNum; i++){
cin >> xx >> yy;
p.push_back(point(xx, yy));
}
}
bool cmp(point a, point b){
if(a.y == b.y) return a.x < b.x;
return a.y < b.y;
}
bool turnRight(point px1, point px2, point pp){
const double eps = 1e-20;
if((px2.x - px1.x)*(pp.y - px2.y) - (pp.x - px2.x)*(px2.y - px1.y) <= eps) return true;
return false;
}
void Compute(){
sort(p.begin(), p.end(), cmp);
CH[0] = 0;
CH[1] = 1;
top = 1;
//從起點0到到排序最後點作凸包右鏈 過程1
for(int i=2; i<nodeNum; i++){
while( top && turnRight(p[CH[top-1]], p[CH[top]], p[i]) )
{
top--;
}
CH[++top] = i;
}
int len = top;
//從排序最高點到到起點0fab反向作凸包右鏈 過程2
CH[++top] = nodeNum-2;
for(int i=nodeNum-3; i>=0; i--){
//top!=len, 不考慮已在過程1生成凸包上的點
while( top!=len && turnRight(p[CH[top-1]], p[CH[top]], p[i]) )
{
top--;
}
CH[++top] = i;
}
}
void Output(){
int sPos;
double Perimeter = 0;
if(cc > 0) cout << endl;
printf("Region #%d:\n", ++cc);
//順時針輸出凸包
for(sPos=0; sPos<top; sPos++){
if(p[CH[sPos]].x == p[0].x && p[CH[sPos]].y == p[0].y)
break;
}
for(int i=top-1; i>=sPos; i--){
if(i!=top-1) printf("-");
printf("(%.1lf,%.1lf)", p[CH[i]].x, p[CH[i]].y);
if(i!=sPos) Perimeter += p[CH[i]].Distance(p[CH[i-1]]);
else Perimeter += p[CH[sPos]].Distance(p[CH[top-1]]);
}
printf("-(%.1lf,%.1lf)\n", p[CH[top-1]].x, p[CH[top-1]].y);
printf("Perimeter length = %.2lf\n", Perimeter);
}
/////////////////////////////////////
int main(){
while(cin >> nodeNum && nodeNum)
{
Input();
Compute();
Output();
}
return 0;
}