HDU - 1392 Surround the Trees (凸包）

Posted 2021-01-06 ckxkexing

Surround the Trees：http://acm.hdu.edu.cn/showproblem.php?pid=1392

题意：

　　在给定点中找到凸包，计算这个凸包的周长。

思路：

　　这道题找出凸包上的点后，s数组中就是按顺序的点，累加一下距离就是周长了。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl ‘
‘

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;       
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 10007;
const double esp = 1e-8;
const double PI=acos(-1.0);



template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar();
    while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
    return x=f?-x:x;
}


/*-----------------------showtime----------------------*/
            const int maxn = 120;
            struct node
            {
                int x,y;
            }p[maxn],s[maxn];
            inline bool cmp(node a,node b){
                double A = atan2((a.y - p[1].y) , (a.x - p[1].x));
                double B = atan2((b.y - p[1].y) , (b.x - p[1].x));
                if(A != B) return A < B;
                else return a.x < b.x;
            }

            ll Cross (node a,node b,node c){
                return 1ll*(b.x - a.x)*(c.y - a.y) - 1ll*(b.y - a.y)*(c.x - a.x);
            }
            int top,n;
            void Get(){
                p[0] = (node){inf,inf}; int k;
                for(int i=1; i<=n; i++)
                    if(p[0].y > p[i].y || (p[0].y == p[i].y && p[i].x < p[0].x))
                    {
                        p[0] = p[i];    k = i;
                    }
                swap(p[k], p[1]);
                sort(&p[2] , &p[n+1], cmp);
                s[0] = p[1] , s[1] = p[2], top = 1;
                for(int i=3; i<=n; ){
                    if(top && Cross(s[top-1] ,p[i], s[top]) >= 0)
                        top--;
                    else s[++top] = p[i++];
                }
            }
            double dis(node a,node b){
                return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));
            }
int main(){     

            while(~scanf("%d", &n) && n){
                for(int i=1; i<=n; i++){
                    scanf("%d%d", &p[i].x, &p[i].y);
                }
                Get();

                double ans = 0;
                if(top == 0){
                    ans = 0;
                }
                else if(top == 1){
                    ans = dis(s[0],s[1]);
                }
                else {
                    s[++top] = s[0];
                    for(int i=0; i<top; i++){
                        ans += dis(s[i], s[i+1]);
                    }
                }
                
                printf("%.2f
", ans);                
            }
            return 0;
}       

HDU - 1392