poj1066 && nyoj83

题目链接 poj1066 & nyoj83

题意：从外面到达宝藏所走的最少的门。
思路：枚举边上的点。

AC代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <stack>#include <queue>#include <vector>#include <cmath>#include <map>#include <set>#define ll long long#define llu unsigned long longusing namespace std;const double eps = 1e-8;struct Point {    double x,y;    Point (double x = 0,double y = 0) : x(x),y(y) {}};Point p[150];struct Line{    Point a,b;};Line line[100];int indl,indp;bool dcmp(double x) {    if(fabs(x)-0.0 <= eps) return true;    return false;}double dis(Point a,Point b) {    return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));}double cross(Point a,Point b,Point c) {    if(dcmp((b.x-a.x)*(c.y-a.y) - (c.x-a.x)*(b.y-a.y))) return 0;    return ((b.x-a.x)*(c.y-a.y) - (c.x-a.x)*(b.y-a.y));}bool f(Point a,Point b,Point c,Point d) {    if(min(a.x,b.x)<=max(c.x,d.x) && min(c.x,d.x)<=max(a.x,b.x)       && min(a.y,b.y)<=max(c.y,d.y) && min(c.y,d.y)<=max(a.y,b.y)) return true;    return false;}int pd(Point a,Point b){    int ans = 0;    for(int i=1;i<indl; i++) {        if(f(a,b,line[i].a,line[i].b)){//        cout<<22222<<endl;        if(cross(a,b,line[i].a)*cross(a,b,line[i].b) <= 0            && cross(line[i].a,line[i].b,a)*(cross(line[i].a,line[i].b,b)) <= 0) {//                cout<<1111<<endl;                ans++;                }        }    }//    cout<<ans<<endl;    return ans;}int main(){    int n;        Point A,B,C;        double a,b,c,d;        scanf("%d",&n);        indl = 1,indp = 1;        for(int i=1;i<=n;i++) {            scanf("%lf%lf%lf%lf",&a,&b,&c,&d);            p[indp].x = a; p[indp].y = b;            indp++;            p[indp].x = c; p[indp].y = d;            indp++;            A.x = a; A.y = b;            B.x = c; B.y = d;            line[indl].a = A;            line[indl].b = B; indl++;        }        scanf("%lf%lf",&C.x,&C.y);        if(n == 0) {            printf("Number of doors = 1\n"); return 0;        }        int Min = 1000000;        for(int i=1;i<indp;i++) {            int f = 0;            for(int j=1; j<indp; j++) {                if(i!=j) {                    if((dcmp(p[i].x-p[j].x)&&dcmp(p[i].y-p[j].y))){                        f = 1; break;                    }                }            }            if(f) continue;//            cout<<i<<"   "<<pd(p[i],C)<<endl;            Min = min(Min,pd(p[i],C));        }        printf("Number of doors = %d\n",Min);    return 0;}