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[題解]TIOJ 1678 剪多邊形(molding)

TIOJ 1678 剪多邊形(molding)

題目連結TIOJ
TIOJ Submission
題目連結ZJ

題目敘述
間單來說是求出多邊形面積以及凸包面積的差,詳細可以點上面題目連結。

題目說多邊形需要才剪下的面積,我們就算凸包面積以及多邊形面積,兩者的差去除上題目給的色塊面積即是答案!

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#include <bits/stdc++.h>
#define Orz ios::sync_with_stdio(0),cin.tie(0)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define pii pair<int,int>
#define pdd pair<double,double>
#define int long long
#define ll long long
#define ld long double
#define N 100001
#define all(x) x.begin(),x.end()
#define eps 1e-9
#define x first
#define y second

using namespace std;

struct pt{
int x,y;
bool operator < (pt b){
if(x == b.x)return y < b.y;
return x < b.x;
}
bool operator > (pt b){
if(x == b.x)return y > b.y;
return x > b.x;
}
bool operator == (pt b){
if(x-b.x == 0 && y-b.y == 0)return true;
return false;
}
pt operator+(pt b) {return {x + b.x, y + b.y};} //向量相加
pt operator-(pt b) {return {x - b.x, y - b.y};} //向量相減
int operator^(pt b) {return x * b.y - y * b.x;} //向量外積cross
int operator*(pt b) {return x * b.x + y * b.y;} //向量內積dot
};

vector<pt> p,temp,pp;
vector<int> cnt;
int n,a,ans = 0;

bool cmp(pt a, pt b){
if(a.x == b.x)return a.y < b.y;
return a.x < b.x;
}

bool check(pt a,pt b,pt o){
pt aa = a - o;
pt bb = b - o;
return (aa^bb) >= 0;
}

vector<pt> solve(){
sort(all(p),cmp);
vector<pt> h;
for(pt i : p){
while(h.size()>=2 && check(i,h[h.size()-1],h[h.size()-2]))
h.pop_back();
h.push_back(i);
}
int sz = h.size();
h.pop_back();
reverse(all(p));
for(auto i : p){
while(h.size()>sz && check(i,h[h.size()-1],h[h.size()-2]))
h.pop_back();
h.push_back(i);
}
return h;
}

int Polygon(){
vector<pt> temp(p);
temp.push_back(temp[0]);
int area = 0;
rep(i,0,n-1)area += (temp[i]^temp[i+1]);
return abs(area);
}

signed main(){
Orz;
cin>>n>>a;
p.resize(n,{0,0});
rep(i,0,n-1)cin>>p[i].x>>p[i].y;
ld small = (ld)Polygon()/2;
vector<pt> hull = solve();
int area = 0,sz = hull.size();
rep(i,0,sz-2)area += (hull[i]^hull[i+1]);
ld big = (ld)area/2;
int ans = ceil((big-small)/a);
cout<<ans<<endl;
}