[poj 1265]Area[Pick定理][三角剖分]

题意:

给出机器人移动的向量, 计算包围区域的内部整点, 边上整点, 面积.

思路:

面积是用三角剖分, 边上整点与GCD有关, 内部整点套用Pick定理.

S = I + E / 2 - 1

I 为内整点数, E为边界整点数, S为面积.

Separate the three numbers by two single blanks.....好吧, 理解成中间空两格PE一次> <

#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int MAXN = 105;
int n;
int GCD(int a, int b)
{
    return !b?a:GCD(b,a%b);
}
struct point
{
    int x,y;
}p[MAXN];
int det(int i, int j)
{
    return p[i].x*p[j].y - p[j].x*p[i].y;
}
double CalS()
{
    double ret = 0;
    for(int i=0;i<n;i++)
    {
        ret += det(i, i+1);
    }
    return fabs(ret/2.0);
}
int CalE()
{
    int ans = n;
    for(int i=0;i<n;i++)
    {
        int dx = (int)abs((double)(p[i].x - p[i+1].x));
        int dy = (int)abs((double)(p[i].y - p[i+1].y));
        if(!dx)
        {
            if(!dy) continue;
            ans += dy - 1;
            continue;
        }
        if(!dx)
        {
            ans += dx - 1;
            continue;
        }
        ans += GCD(dx, dy) - 1;
    }
    return ans;
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int k=1;k<=T;k++)
    {
        scanf("%d",&n);
        p[0].x = p[0].y = 0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d %d",&p[i].x,&p[i].y);
            p[i].x += p[i-1].x, p[i].y += p[i-1].y;
        }
        double S = CalS();
        int E = CalE();
        printf("Scenario #%d:
%d %d %.1lf

",k,(int)(S+1.0-E/2.0),E,S);
    }
}