[ZOJ1123] Triangle Encapsulation

題目大意：

順時針給出一個三角形的三個頂點，按照題中格式打印出所有三角形內（邊上的不算）的整數座標點。

解題思路：

枚舉座標範圍內左右的點，計算點對三角形每條邊兩個頂點的夾角（若對某條邊兩個頂點的夾角爲PI，則在邊上），求夾角和，若夾角和爲2*PI，則點在三角形內。

注意點：

Sample Output裏的Program 4 by team X和End of program 4 by team X要打印。

源代碼

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define maxn 30

const double PI=acos(-1.0), EPS=1e-6, PI2=2*acos(-1.0);

int x[4], y[4];

double dbcmp(double a, double b)
{
    return b-a>EPS?-1:a-b>EPS;
}

double radian(int ax, int ay, int bx, int by)
{
    double r=atan2(1.0*ax*by-(ay*bx), 1.0*ax*bx+(ay*by));
    return r;
}

bool in_tri(const int xx, const int yy)
{
    double s=0, t=0;
    for (int i=0; i<3; i++)
    {
        t=radian(x[i+1]-xx, y[i+1]-yy, x[i]-xx, y[i]-yy);
        if (dbcmp(t, PI)==0) return false;
        s+=t;
    }
    if (dbcmp(s, PI2)==0) return true;
    return false;
}

int min(int xx, int yy){if (xx<yy) return xx; return yy;}
int max(int xx, int yy){if (xx>yy) return xx; return yy;}

int main()
{
//    freopen("test.txt", "r", stdin);
//    freopen("ans.txt", "w", stdout);
    printf("Program 4 by team X\n");
    int  minx, maxx, miny, maxy, ansminx;
    int end_x[maxn];            //last dot in line i end with (end_x[i], i)
    bool ans[maxn][maxn];
    while (scanf("%d%d%d%d%d%d", &x[0],&y[0],&x[1],&y[1],&x[2],&y[2])==6)
    { 
        memset(ans, 0, sizeof(ans));
        memset(end_x, 255, sizeof(end_x));
        minx=miny=maxn+1;
        maxx=maxy=-1;
        for (int i=0; i<3; i++)
        {
            x[i]+=10; y[i]+=10;
            minx=min(minx, x[i]); maxx=max(maxx, x[i]);
            miny=min(miny, y[i]); maxy=max(maxy, y[i]);
        }
        x[3]=x[0]; y[3]=y[0]; ansminx=maxn+1;
        for (int xx=minx; xx<=maxx; xx++)
            for (int yy=miny; yy<=maxy; yy++)
                if (in_tri(xx,yy)){
                    ans[xx][yy]=1;
                    if (end_x[yy]==-1 || xx>end_x[yy]) end_x[yy]=xx;
                    ansminx=min(xx, ansminx);
                }
        for (int yy=maxy; yy>=0; yy--)
            if (end_x[yy]!=-1)
            {
                for (int xx=ansminx; xx<end_x[yy]; xx++)
                    if (ans[xx][yy]) printf("(%2d, %2d) ", xx-10, yy-10);
                    else printf("         ");
                if (ans[end_x[yy]][yy]) printf("(%2d, %2d)\n", end_x[yy]-10, yy-10);
            }
        printf("\n");
    }
    printf("End of program 4 by team X\n");
    return 0;
}