POJ2318 TOYS(叉积),POJ2398 TOY STORAGE

题意,给出一堆线段和点,求线段分割的区域中点的个数.

计算几何基础题的模板差不多也搞出来了...

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <cmath>
#include <algorithm>
const double EPS = 1e-8;
using namespace std;
/********************************/

int dcmp(double x){
if(fabs(x) < EPS)
return 0;
if(x < 0) return -1;
return 1;//1为正，-1为负，0为等
}
struct point{
double x , y;
point(double x = 0,double y= 0):x(x),y(y){}

/************************************/
};
typedef point Vector;
point operator + (const Vector &a , const Vector &b){
return Vector(a.x + b.x,a.y + b.y);
}
point operator - (const Vector &a , const Vector &b){
return Vector(a.x - b.x,a.y - b.y);
}
point operator * (const Vector &a , const double &p){
return Vector(a.x * p,a.y * p);
}
point operator / (const Vector &a , const double &p){
return Vector(a.x / p,a.y / p);
}
bool operator < (const point &a,const point &b){
return a.x < b.x || (a.x == b.x && a.y < b.y);
}
bool operator == (const point &a,const point &b){
return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) ==0;
};
double cross(const Vector &a ,const Vector &b)
{
return (a.x * b.y - a.y * b.x);
}
point edge[6000][2];
int cnt[6000];
int main()
{
//freopen("in.txt","r",stdin);
int n , m;
int x1 ,x2 ,y1, y2 , Ui,Li;
while(scanf("%d",&n) , n){

memset(cnt,0,sizeof(cnt));
if(n == 0)
return 0;

scanf("%d%d%d%d%d" , &m,&x1,&y1,&x2,&y2);
edge[0][0] = point(x1 , y1);
edge[0][1] = point(x1 , y2);
for(int i = 1 ;i <= n ; i++){
scanf("%d%d",&Ui,&Li);
edge[i][0] = point(Ui , y1);
edge[i][1] = point(Li , y2);
}
edge[n+1][0] = point(x2,y1);
edge[n+1][1] = point(x2,y2);
int findx,findy;
point d;
for(int c = 0 ;c < m ; c++){
scanf("%d%d" , &findx , &findy);
d = point(findx,findy);
for(int i = 1; i <= n+1 ;i++){
if(cross(edge[i][1] - edge[i][0] , d - edge[i][0]) <0)
{
cnt[i-1]++;
break;
}
}
}
for(int i = 0 ;i <= n; i++)
printf("%d: %d\n",i,cnt[i]);
printf("\n");
}
return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <cmath>
#include <algorithm>
const double EPS = 1e-8;
using namespace std;
/********************************/

int dcmp(double x){
if(fabs(x) < EPS)
return 0;
if(x < 0) return -1;
return 1;//1为正，-1为负，0为等
}
struct point{
double x , y;
point(double x = 0,double y= 0):x(x),y(y){}

/************************************/
};
typedef point Vector;
point operator + (const Vector &a , const Vector &b){
return Vector(a.x + b.x,a.y + b.y);
}
point operator - (const Vector &a , const Vector &b){
return Vector(a.x - b.x,a.y - b.y);
}
point operator * (const Vector &a , const double &p){
return Vector(a.x * p,a.y * p);
}
point operator / (const Vector &a , const double &p){
return Vector(a.x / p,a.y / p);
}
bool operator > (const point &a,const point &b){
return a.x > b.x;
}
bool operator == (const point &a,const point &b){
return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) ==0;
};
double cross(const Vector &a ,const Vector &b)
{
return (a.x * b.y - a.y * b.x);
}
point edge[2000][2];
int cnt[2000];
int res[2000];
int main()
{
//freopen("in.txt","r",stdin);
int n , m;
int x1 ,x2 ,y1, y2 , Ui,Li;
while(scanf("%d",&n) , n){

memset(cnt,0,sizeof(cnt));
memset(res,0,sizeof(res));
if(n == 0)
return 0;

scanf("%d%d%d%d%d" , &m,&x1,&y1,&x2,&y2);

edge[0][0] = point(x1 , y1);
edge[0][1] = point(x1 , y2);
for(int i = 1 ;i <= n ; i++){
scanf("%d%d",&Ui,&Li);
edge[i][0] = point(Ui , y1);
edge[i][1] = point(Li , y2);
}
edge[n+1][0] = point(x2,y1);
edge[n+1][1] = point(x2,y2);

for(int i = 1 ; i <= n;i++)
for(int j = 1 ; j<= n-i ; j++)
if(edge[j][0] > edge[j+1][0]){
swap(edge[j][0],edge[j+1][0]);
swap(edge[j][1],edge[j+1][1]);
}

int findx,findy;
point d;
for(int c = 0 ;c < m ; c++){
scanf("%d%d" , &findx , &findy);
d = point(findx,findy);
for(int i = 1; i <= n+1 ;i++){
if(cross(edge[i][1] - edge[i][0] , d - edge[i][0]) <0)
{
cnt[i-1]++;
break;
}
}
}
printf("Box\n");
for(int i = 0 ;i <= n;i++)
res[cnt[i]]++;
for(int i= 1 ;i < 1999;i++)
if(res[i] != 0)
printf("%d: %d\n",i,res[i]);

}
return 0;
}