Red Huang

Red Huang

uva 143

利用面積求法即可,記得精度問題

還有如果三點共線也必須找出在那條線上的頂點

//  
//        GGGGGGGGGGGGG        CCCCCCCCCCCCC               AAA  
//     GGG::::::::::::G     CCC::::::::::::C              A:::A  
//   GG:::::::::::::::G   CC:::::::::::::::C             A:::::A  
//  G:::::GGGGGGGG::::G  C:::::CCCCCCCC::::C            A:::::::A  
// G:::::G       GGGGGG C:::::C       CCCCCC           A:::::::::A  
//G:::::G              C:::::C                        A:::::A:::::A  
//G:::::G              C:::::C                       A:::::A A:::::A  
//G:::::G    GGGGGGGGGGC:::::C                      A:::::A   A:::::A  
//G:::::G    G::::::::GC:::::C                     A:::::A     A:::::A  
//G:::::G    GGGGG::::GC:::::C                    A:::::AAAAAAAAA:::::A  
//G:::::G        G::::GC:::::C                   A:::::::::::::::::::::A  
// G:::::G       G::::G C:::::C       CCCCCC    A:::::AAAAAAAAAAAAA:::::A  
//  G:::::GGGGGGGG::::G  C:::::CCCCCCCC::::C   A:::::A             A:::::A  
//   GG:::::::::::::::G   CC:::::::::::::::C  A:::::A               A:::::A  
//     GGG::::::GGG:::G     CCC::::::::::::C A:::::A                 A:::::A  
//        GGGGGG   GGGG        CCCCCCCCCCCCCAAAAAAA                   AAAAAAA  
#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <algorithm>  
#include <cmath>  
#include <climits>  
#include <vector>  
#include <set>  
#include <map>  
#include <queue>  
#include <cctype>  
#include <utility>  
#include <ctime>  
using namespace std;  
#ifdef DEBUG  
#define debug(...) printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)  
#define gettime() end\_time=clock();printf("now running time is %.7f\\n",(float)(end\_time - start\_time)/CLOCKS\_PER\_SEC);  
#else  
#define debug(...)  
#define gettime()  
#endif  
typedef unsigned int uint;  
typedef long long int Int;  
#define Set(a,s) memset(a,s,sizeof(a))  
#define Write(w) freopen(w,"w",stdout)  
#define Read(r) freopen(r,"r",stdin)  
#define Pln() printf("\\n")  
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()  
#define De(x)printf(#x"%d\\n",x)  
#define For(i,x)for(int i=0;i<x;i++)  
#define CON(x,y) x##y  
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\  
    printf("%d ",dp\[htx\]\[hty\]);}Pln();}  
#define M 55  
#define PII pair<int,int\>  
#define PB push\_back  
#define oo INT\_MAX  
#define Set\_oo 0x3f  
#define FOR(it,c) for(vector<PII>::iterator it=(c).begin();it!=(c).end();it++)  
#define eps 1e-6  
clock\_t start\_time=clock(), end\_time;  
bool xdy(double x,double y){return x>y+eps;}  
bool xddy(double x,double y){return x>y-eps;}  
bool xcy(double x,double y){return x<y-eps;}  
bool xcdy(double x,double y){return x<y+eps;}  
bool xeqy(double x,double y){return fabs(x-y)<eps;}  
int min3(int x,int y,int z){  
    int tmp=min(x,y);  
    return min(tmp,z);  
}  
int max3(int x,int y,int z){  
    int tmp=max(x,y);  
    return max(tmp,z);  
}  
struct points{  
    double x,y;  
}p\[3\];  
double maxx,maxy,minx,miny;  
double cal(double x1,double y1,double x2,double y2){  
    return x1\*y2-x2\*y1;  
}  
bool isinside(points o){  
    double area=0;  
    double tarea=fabs(cal(p\[1\].x-p\[0\].x,p\[1\].y-p\[0\].y,p\[2\].x-p\[0\].x,p\[2\].y-p\[0\].y));  
    for(int i=0;i<3;i++){  
        int j=(i+1)%3;  
        area+=fabs(cal(p\[i\].x-o.x,p\[i\].y-o.y,p\[j\].x-o.x,p\[j\].y-o.y));  
//        debug("%.30f %.10f\\n",area,tarea);  
    }  
    if(xeqy(area,tarea)){  
        return true;  
    }  
    return false;  
}  
void solve(){  
    points o;  
    int ans=0;  
    for(int i=ceil(minx);i<=floor(maxx);i++){  
        for(int j=ceil(miny);j<=floor(maxy);j++){  
//            debug("%d %d\\n",i,j);  
            o.x=i;o.y=j;  
            if(isinside(o))ans++;  
        }  
    }  
    printf("%4d\\n",ans);  
}  
int main() {  
    ios\_base::sync\_with\_stdio(0);  
    double all=0;  
    while(~scanf("%lf",&p\[0\].x)){  
        scanf("%lf",&p\[0\].y);  
        minx=maxx=p\[0\].x;miny=maxy=p\[0\].y;  
        all=p\[0\].x+p\[0\].y;  
        for(int i=1;i<3;i++){  
            scanf("%lf%lf",&p\[i\].x,&p\[i\].y);  
            minx=min(minx,p\[i\].x);  
            maxx=max(maxx,p\[i\].x);  
            miny=min(miny,p\[i\].y);  
            maxy=max(maxy,p\[i\].y);  
            all+=p\[i\].x+p\[i\].y;  
        }  
        if(all==0)break;  
        minx=max(minx,1.0);  
        miny=max(miny,1.0);  
        maxx=min(maxx,99.0);  
        maxy=min(maxy,99.0);  
        solve();  
    }  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
}  

加载中...
此文章数据所有权由区块链加密技术和智能合约保障仅归创作者所有。