poj1066Treasure Hunt（线段相交）

链接

很纠结的找到了所有线段的中点，又很纠结的找到了哪些中点可以直接相连，最后bfs一下求出了最短路。。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 2210
#define LL long long
#define INF 0xfffffff
const double eps = 1e-10;
const double inf = ~0u>>2;
vector<int>ed[N];
struct Point
{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y) {} //构造函数 方便代码编写
} p[105],q[N];
int g,dis[N];
bool vis[N];
vector<Point>pi[N];
typedef Point pointt;
pointt operator + (Point a,Point b)
{
    return Point(a.x+b.x,a.y+b.y);
}
pointt operator - (Point a,Point b)
{
    return Point(a.x-b.x,a.y-b.y);
}
pointt operator * (Point a,double b)
{
    return Point(a.x*b,a.y*b);
}
pointt operator / (Point a,double b)
{
    return Point(a.x/b,a.y/b);
}
bool operator < (const Point &a,const Point &b)
{
    return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    else return x<0?-1:1;
}
double cross(Point a,Point b)
{
    return a.x*b.y-a.y*b.x;
}
Point intersection1(Point a,Point b,Point c,Point d)//直线交点求解
{
    Point pp = a;
    double t = ((a.x-c.x)*(c.y-d.y)-(a.y-c.y)*(c.x-d.x))/
               ((a.x-b.x)*(c.y-d.y)-(a.y-b.y)*(c.x-d.x));
    pp.x+=(b.x-a.x)*t;
    pp.y+=(b.y-a.y)*t;
    return pp;
}
bool seginter(pointt a1,pointt a2,pointt b1,pointt b2)
{
    double c1 = cross(a2-a1,b1-a1),c2 = cross(a2-a1,b2-a1),
                                        c3 = cross(b2-b1,a1-b1),c4 = cross(b2-b1,a2-b1);
    return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}
void init(int i,int n)
{
    p[i+1].x = 0;
    p[i+1].y = 0;
    p[i+2].x = 0;
    p[i+2].y = 100;
    p[i+3].x = 100;
    p[i+3].y = 100;
    p[i+4].x = 100;
    p[i+4].y = 0;
    p[i+1+n] = p[i+2];
    p[i+2+n] = p[i+3];
    p[i+3+n] = p[i+4];
    p[i+4+n] = p[i+1];

}
int judge(int x)
{
    if(fabs(q[x].x)<eps||fabs(q[x].y)<eps||fabs(q[x].x-100.0)<eps||fabs(q[x].y-100.0)<eps)
        return 1;
    return 0;
}
void bfs(int st)
{
    queue<int>Q;
    Q.push(st);
    memset(vis,0,sizeof(vis));
   // cut<<
    for(int i = 0 ; i <= st ; i++)
        dis[i] = INF;
    dis[st] = 0;
    while(!Q.empty())
    {
        int u = Q.front();
        Q.pop();
        //printf("%.2f %.2f %d
",q[u].x,q[u].y,dis[u]);
        if(judge(u))
        {
            printf("Number of doors = %d
",dis[u]);
            break;
        }
        for(int i =0 ; i < ed[u].size() ; i++)
        {
            int v = ed[u][i];
            //cout<<v<<" "<<dis[u]<<endl;
            dis[v] = min(dis[v],dis[u]+1);
            if(!vis[v])
            {
                vis[v] = 1;
                Q.push(v);
            }
        }
    }
}
bool cmp(Point a,Point b)
{
    if(fabs(a.x-b.x)<eps)
        return a.y<b.y;
    return a.x<b.x;
}
int main()
{
    int n,i,j,e;
    scanf("%d",&n);
//    while(scanf("%d",&n)!=EOF)
//    {
        n+=4;
        g = 0;
        for(i = 1; i <= n-4 ; i++)
        {
            scanf("%lf%lf%lf%lf",&p[i].x,&p[i].y,&p[i+n].x,&p[i+n].y);
        }
        init(n-4,n);
        for(i = 1; i <= n-4; i++)
        {
            for(j = 1; j <= n-4 ; j++)
            {
                if(i==j) continue;
                if(seginter(p[i],p[i+n],p[j],p[j+n]))
                {
                    pi[i].push_back(intersection1(p[i],p[i+n],p[j],p[j+n]));
                }
            }
        }
        for(i = 1 ; i <= n-4 ; i++)
        {
            if(dcmp(p[i].x)==0)
                pi[n-3].push_back(p[i]);
            if(dcmp(p[i+n].x)==0)
                pi[n-3].push_back(p[i+n]);
            if(dcmp(p[i].y-100)==0)
                pi[n-2].push_back(p[i]);
            if(dcmp(p[i+n].y-100)==0)
                pi[n-2].push_back(p[i+n]);
            if(dcmp(p[i].x-100)==0)
                pi[n-1].push_back(p[i]);
            if(dcmp(p[i+n].x-100)==0)
                pi[n-1].push_back(p[i+n]);
            if(dcmp(p[i].y)==0)
                pi[n].push_back(p[i]);
            if(dcmp(p[i+n].y)==0)
                pi[n].push_back(p[i+n]);
        }
        for(i = 1 ; i <= n; i++)
        {
            pi[i].push_back(p[i]);
            pi[i].push_back(p[i+n]);
            sort(pi[i].begin(),pi[i].end(),cmp);
            for(j = 0 ; j < pi[i].size()-1 ; j++)
            {
                Point tp;
                Point p1 = pi[i][j],p2= pi[i][j+1];
                if(dcmp(p1.x-p2.x)==0&&dcmp(p1.y-p2.y)==0) continue;
                tp.x = (p1.x+p2.x)/2;
                tp.y = (p1.y+p2.y)/2;
                // printf("%d tpx = %.2f tpy = %.2f p1x = %.2f p1y = %.2f p2x = %.2f p2y = %.2f
",i,tp.x,tp.y,p1.x,p1.y,p2.x,p2.y);
                q[++g] = tp;
            }
            pi[i].clear();
        }
        g++;
        scanf("%lf%lf",&q[g].x,&q[g].y);
        for(i = 1; i <= g ; i++)
            ed[i].clear();
        for(i = 1; i <= g ; i++)
        {
            //  printf("%.2f %.2f
",q[i].x,q[i].y);
            for(j = 1; j <= g ; j++)
            {
                if(i==j) continue;
                for(e = 1; e <= n; e++)
                {
                    if(seginter(q[i],q[j],p[e],p[e+n]))
                    {
//                        if(i==g)
//                        printf("%.2f %.2f e = %d
",q[j].x,q[j].y,e);
                        break;
                    }
                }
                if(e>n)
                {
//                    if(i==g)
//                    printf("%.2f %.2f %.2f %.2f
",q[i].x,q[i].y,q[j].x,q[j].y);
                    ed[i].push_back(j);
                }
            }
        }
        bfs(g);
//    }
    return 0;
}

感觉这题没那么复杂，交对后搜了下题解，，然后 这题是一个水题，确实水。。

因为这个点的最终目的是要走出去的，边界上墙的点相对于边界上其它的点应该是更优的，所以只需枚举这些点到达起点经过多少堵墙即可。

几何抓不住重点是件很纠结的事。

附优美的简短的跑得飞快的代码

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 2210
#define LL long long
#define INF 0xfffffff
const double eps = 1e-10;
const double inf = ~0u>>2;
vector<int>ed[N];
struct Point
{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y) {} //构造函数 方便代码编写
} p[105],q[N];
int g,dis[N];
bool vis[N];
vector<Point>pi[N];
typedef Point pointt;
pointt operator + (Point a,Point b)
{
    return Point(a.x+b.x,a.y+b.y);
}
pointt operator - (Point a,Point b)
{
    return Point(a.x-b.x,a.y-b.y);
}
pointt operator * (Point a,double b)
{
    return Point(a.x*b,a.y*b);
}
pointt operator / (Point a,double b)
{
    return Point(a.x/b,a.y/b);
}
bool operator < (const Point &a,const Point &b)
{
    return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    else return x<0?-1:1;
}
double cross(Point a,Point b)
{
    return a.x*b.y-a.y*b.x;
}
bool seginter(pointt a1,pointt a2,pointt b1,pointt b2)
{
    double c1 = cross(a2-a1,b1-a1),c2 = cross(a2-a1,b2-a1),
                                        c3 = cross(b2-b1,a1-b1),c4 = cross(b2-b1,a2-b1);
    return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}
int main()
{
    int n,i,j,e;
    scanf("%d",&n);
    g = 0;
    for(i = 1; i <= n ; i++)
    {
        scanf("%lf%lf%lf%lf",&p[i].x,&p[i].y,&p[i+n].x,&p[i+n].y);
    }
    Point tp;
    scanf("%lf%lf",&tp.x,&tp.y);
    if(n==0)
    {
        puts("Number of doors = 1");
        return 0;
    }
    int ans = INF;
    for(i = 1 ; i <= n*2; i++)
    {
        int ts = 1;
        for(j = 1 ; j <= n ; j++)
        {
            if(seginter(p[i],tp,p[j],p[j+n]))
                ts++;
        }
        ans = min(ans,ts);
    }
    printf("Number of doors = %d
",ans);
    return 0;
}