凸包
POJ 1113
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long ll;
const double eps = 1e-9;
const double PI = acos(-1.0);
inline int sgn(double x) {
if (x < -eps) return -1;
if (x > eps) return 1;
return 0;
}
struct vec {
double x, y;
vec() {x = y = 0;}
vec(double _x, double _y): x(_x), y(_y) {}
vec operator + (vec v) { return vec(x + v.x, y + v.y);}
vec operator - (vec v) { return vec(x - v.x, y - v.y);}
vec operator * (double v) {return vec(x * v, y * v);}
vec operator / (double v) {return vec(x / v, y / v);}
double operator * (vec v) {return x * v.x + y * v.y;}
double len() {return hypot(x, y);}
double lensqr() {return x * x + y * y;}
} V[1005], Z[1005];
double cross(vec a, vec b) {
return a.x * b.y - a.y * b.x;
}
double dist(double x1, double y1, double x2, double y2) {
return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
bool cmpXY(vec a, vec b) {
if (sgn(a.x - b.x)) return a.x < b.x;
return a.y < b.y;
}
double convex_hull(vec* v, int n, vec *z) {
sort(v, v + n, cmpXY);
int m = 0;
for (int i = 0; i < n; i++) {
while (m > 1 && cross(z[m - 1] - z[m - 2], v[i] - z[m - 2]) <= 0) --m;
z[m++] = v[i];
}
int k = m;
for (int i = n - 2; i >= 0; i--) {
while (m > k && cross(z[m - 1] - z[m - 2], v[i] - z[m - 2]) <= 0) --m;
z[m++] = v[i];
}
if (n > 1) --m;
double ret = 0;
for (int i = 0; i < m; i++)
ret += dist(z[i].x, z[i].y, z[(i + 1) % m].x, z[(i + 1) % m].y);
return ret;
}
int main() {
//freopen("in.txt","r",stdin);
int n;
double r;
while (~scanf("%d", &n)) {
scanf("%lf", &r);
for (int i = 0; i < n; i++)
scanf("%lf%lf", &V[i].x, &V[i].y);
double ret = convex_hull(V, n, Z);
ret += PI * r * 2.0;
int ans = ret + 0.5;
printf("%d
", ans);
}
return 0;
}
BZOJ 1185
#include<cstdio>
#include<cmath>
#include<ctime>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<set>
#define eps 1e-8
#define inf 1000000000
using namespace std;
double ans=1e60;
int n,top;
struct P{
double x,y;
P(){}
P(double _x,double _y):x(_x),y(_y){}
friend bool operator<(P a,P b){
return fabs(a.y-b.y)<eps?a.x<b.x:a.y<b.y;
}
friend bool operator==(P a,P b){
return fabs(a.x-b.x)<eps&&fabs(a.y-b.y)<eps;
}
friend bool operator!=(P a,P b){
return !(a==b);
}
friend P operator+(P a,P b){
return P(a.x+b.x,a.y+b.y);
}
friend P operator-(P a,P b){
return P(a.x-b.x,a.y-b.y);
}
friend double operator*(P a,P b){
return a.x*b.y-a.y*b.x;
}
friend P operator*(P a,double b){
return P(a.x*b,a.y*b);
}
friend double operator/(P a,P b){
return a.x*b.x+a.y*b.y;
}
friend double dis(P a){
return sqrt(a.x*a.x+a.y*a.y);
}
}p[50005],q[50005],t[5];
bool cmp(P a,P b)
{
double t=(a-p[1])*(b-p[1]);
if(fabs(t)<eps)return dis(p[1]-a)-dis(p[1]-b)<0;
return t>0;
}
void graham()
{
for(int i=2;i<=n;i++)
if(p[i]<p[1])
swap(p[i],p[1]);
sort(p+2,p+n+1,cmp);
q[++top]=p[1];
for(int i=2;i<=n;i++)
{
while(top>1&&(q[top]-q[top-1])*(p[i]-q[top])<eps)top--;
q[++top]=p[i];
}
q[0]=q[top];
}
void RC()
{
int l=1,r=1,p=1;
double L,R,D,H;
for(int i=0;i<top;i++)
{
D=dis(q[i]-q[i+1]);
while((q[i+1]-q[i])*(q[p+1]-q[i])-(q[i+1]-q[i])*(q[p]-q[i])>-eps)p=(p+1)%top;
while((q[i+1]-q[i])/(q[r+1]-q[i])-(q[i+1]-q[i])/(q[r]-q[i])>-eps)r=(r+1)%top;
if(i==0)l=r;
while((q[i+1]-q[i])/(q[l+1]-q[i])-(q[i+1]-q[i])/(q[l]-q[i])<eps)l=(l+1)%top;
L=(q[i+1]-q[i])/(q[l]-q[i])/D,R=(q[i+1]-q[i])/(q[r]-q[i])/D;
H=(q[i+1]-q[i])*(q[p]-q[i])/D;
if(H<0)H=-H;
double tmp=(R-L)*H;
if(tmp<ans)
{
ans=tmp;
t[0]=q[i]+(q[i+1]-q[i])*(R/D);
t[1]=t[0]+(q[r]-t[0])*(H/dis(t[0]-q[r]));
t[2]=t[1]-(t[0]-q[i])*((R-L)/dis(q[i]-t[0]));
t[3]=t[2]-(t[1]-t[0]);
}
}
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
graham();
RC();
printf("%.5lf
",ans);
int fir=0;
for(int i=1;i<=3;i++)
if(t[i]<t[fir])
fir=i;
for(int i=0;i<=3;i++)
printf("%.5lf %.5lf
",t[(i+fir)%4].x,t[(i+fir)%4].y);
return 0;
}
最大空凸包
POJ 1259 HDU 6219
dp[i][j]表示以<i,j>为以o为原点选的最后一个三角形的边。(i>j)
需要确定三角形内无点
那么dp[i][j]=dp[j][k]+S(i,j,o)。需要<j,i>与<k,j>形成凸边。
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long ll;
const double eps = 1e-9;
const double PI = acos(-1.0);
const int maxn = 405;
double dp[maxn][maxn];
int T,n;
inline int sgn(double x) {
if (x < -eps) return -1;
if (x > eps) return 1;
return 0;
}
struct vec {
double x, y;
vec() {x = y = 0;}
vec(double _x, double _y): x(_x), y(_y) {}
vec operator + (vec v) { return vec(x + v.x, y + v.y);}
vec operator - (vec v) { return vec(x - v.x, y - v.y);}
vec operator * (double v) {return vec(x * v, y * v);}
vec operator / (double v) {return vec(x / v, y / v);}
double operator * (vec v) {return x * v.x + y * v.y;}
double len() {return hypot(x, y);}
double lensqr() {return x * x + y * y;}
} V[maxn], Z[maxn],o;
double cross(vec a, vec b) {
return a.x * b.y - a.y * b.x;
}
double dist(double x1, double y1, double x2, double y2) {
return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
bool cmpangle(vec a,vec b){
int k=sgn(cross(a-o,b-o));
if(k==0){
return (a-o).lensqr()<(b-o).lensqr();
//return dist(a.x,a.y,o.x,o.y)<dist(b.x,b.y,o.x,o.y);
} else return k>0;
}
double calc_empty_convex(vec Z[],int n){
for(int i=0;i<=n;i++)
for(int j=0;j<=n;j++)
dp[i][j]=0;
double ret=0;
int j,k;
for(int i=0;i<n;i++){
j=i-1;
while(j>=0 && sgn(cross(Z[i]-o,Z[j]-o))==0) j--;
bool flg=(j==i-1);
while(j>=0){
k=j-1;
while(k>=0&&cross(Z[i]-Z[j],Z[j]-Z[k])>0) k--;
double area=fabs(cross(Z[i]-o,Z[j]-o))*0.5;
if(k>=0) area+=dp[j][k];
// cout<<area<<endl;
if(flg) dp[i][j]=area;
ret=max(area,ret);
j=k;
}
if(flg){
for(int c=1;c<i;c++)
dp[i][c]=max(dp[i][c],dp[i][c-1]);
}
}
//cout<<ret<<endl;
return ret;
}
double empty_convex(vec V[],int n){
double ret=0;
for(int i=0;i<n;i++){
o=V[i];
int top=0;
for(int j=0;j<n;j++)
if(V[j].y>o.y||(sgn(V[j].y-o.y)==0&&sgn(V[j].x-o.x)>=0)){
Z[top++]=V[j];
}
sort(Z,Z+top,cmpangle);
ret=max(ret,calc_empty_convex(Z,top));
}
return ret;
}
int main() {
//freopen("in.txt","r",stdin);
cin>>T;
while(T--){
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%lf%lf",&V[i].x,&V[i].y);
double res = empty_convex(V,n);
printf("%.1f
",res);
}
return 0;
}