HDU 4380 Farmer Greedy (计算几何)

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传送门：http://acm.hdu.edu.cn/showproblem.php?pid=4380

题意：给你n个房子的坐标，给你m个金子的坐标，问你有多少个用房子围成的三角形里面包含奇数个金子。

题解：求n个点任意两个点连成的线段下面有多少个金子，然后3个线段相减判断是不是奇数就行了。

注意一点，判断线段下面多少个点时用到两个点相乘，可能超过int，所以用__int64。

这里被freopen("input.txt","r",stdin);坑爹了，忘记注释了，wa了四次……

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <cmath>#include <vector>#include <list>#include <deque>#include <queue>#include <iterator>#include <stack>#include <map>#include <set>#include <algorithm>#include <cctype>using namespace std;#define si1(a) scanf("%d",&a)#define si2(a,b) scanf("%d%d",&a,&b)#define sd1(a) scanf("%lf",&a)#define sd2(a,b) scanf("%lf%lf",&a,&b)#define ss1(s)  scanf("%s",s)#define pi1(a)    printf("%d\n",a)#define pi2(a,b)  printf("%d %d\n",a,b)#define mset(a,b)   memset(a,b,sizeof(a))#define forb(i,a,b)   for(int i=a;i<b;i++)#define ford(i,a,b)   for(int i=a;i<=b;i++)typedef __int64 LL;const int N=110;const int mod=1000007;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);const double eps=1e-7;int n,m;int xh[N][N];   //xh[i][j]表示线段ij下面有多少个点struct point{    LL x,y;    void input()    {        scanf("%I64d%I64d",&x,&y);    }}house[110],gold[1010];bool cmp(point a,point b){    if(a.x!=b.x)        return a.x<b.x;    return a.y<b.y;}bool chacheng(point a,point b,point c){    point u,v;    u.x=b.x-a.x;    u.y=b.y-a.y;    v.x=c.x-b.x;    v.y=c.y-b.y;    LL t=u.x*v.y-u.y*v.x;    if(t>=0)        return true;    return false;}bool under(int i,int j,int k){    if(gold[k].x<house[i].x||gold[k].x>=house[j].x)        return false;    return chacheng(house[j],house[i],gold[k]);}int main(){//    freopen("input.txt","r",stdin);    int ca=0;    while(si2(n,m)!=EOF)    {        forb(i,0,n)            house[i].input();        forb(i,0,m)            gold[i].input();        sort(house,house+n,cmp);        mset(xh,0);        forb(i,0,n)            forb(j,i+1,n)                forb(k,0,m)                    if(under(i,j,k))                        xh[i][j]++,xh[j][i]++;        int num=0;        forb(i,0,n)            forb(j,i+1,n)                forb(k,j+1,n)                    if(abs(xh[i][k]-xh[i][j]-xh[j][k])&1)                        num++;        printf("Case %d: %d\n",++ca,num);    }    return 0;}