poj1066 && nyoj83

最新推荐文章于 2017-11-22 18:43:00 发布

原创最新推荐文章于 2017-11-22 18:43:00 发布 · 255 阅读

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本文针对POJ1066和NYOJ83题目进行了解析，该问题是寻找通往宝藏路径上需要经过的最少门数。通过枚举边上的点并使用点线距离判断来实现解决方案。

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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 long
using 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;
}