POJ 1556 The Doors（建图+最短路）

蒻苣：弱弱很弱，路过的巨巨还不吝赐教！^.^...QAQ

题意：给你n<18堵墙，每个墙输入五个点，有两扇门，求从(0,5)->(10,5)的最短路

题解：枚举每个点，到其它点不与直线相交的边（也就是构图），然而最短路一发即可.

代码：

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
using namespace std;
const double INF  = 1E200;
const double wc = 1E-8;
struct POINT{
        double x,y;
        POINT( double a = 0,double b = 0 )  { x = a; y = b; }
};
POINT t[10000];
struct LINESEG{
        POINT e;
        POINT s;
        LINESEG( POINT x,POINT y ) { e = x; s = y; }
        LINESEG(){}
};
LINESEG L[1000];
double dist(POINT p1,POINT p2)                // 返回两点之间欧氏距离
{
 return( sqrt( (p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y) ) );
}
double multiply(POINT sp,POINT ep,POINT op)
{
 return((sp.x-op.x)*(ep.y-op.y)-(ep.x-op.x)*(sp.y-op.y));
}
bool online(LINESEG l,POINT p)
{
 return( (multiply(l.e,p,l.s)==0) &&( ( (p.x-l.s.x)*(p.x-l.e.x)<=0 )&&( (p.y-l.s.y)*(p.y-l.e.y)<=0 ) ) );
}
bool intersect(LINESEG u,LINESEG v)
{
 return( (max(u.s.x,u.e.x)>=min(v.s.x,v.e.x))&&                     //排斥实验
   (max(v.s.x,v.e.x)>=min(u.s.x,u.e.x))&&
   (max(u.s.y,u.e.y)>=min(v.s.y,v.e.y))&&
   (max(v.s.y,v.e.y)>=min(u.s.y,u.e.y))&&
   (multiply(v.s,u.e,u.s)*multiply(u.e,v.e,u.s)>=wc)&&         //跨立实验
   (multiply(u.s,v.e,v.s)*multiply(v.e,u.e,v.s)>=wc));
}
//  (线段u和v相交)&&(交点不是双方的端点) 时返回true
bool intersect_A(LINESEG u,LINESEG v)
{
 return ((intersect(u,v))&&
   (!online(u,v.s))&&
   (!online(u,v.e))&&
   (!online(v,u.e))&&
   (!online(v,u.s)));
}
bool intersect_l(LINESEG u,LINESEG v)
{
 return multiply(u.s,v.e,v.s)*multiply(v.e,u.e,v.s)>=wc;
}
//----------------
//使用优先队列优化：时间复杂度为O(E*logE),E为定点个数
//
//注意对vector<Edge>E[MAXN]进行初始化加边
//
//------------------
const int MAXN = 10010;
struct qnode
{
    int v;
    double c;
    qnode(int _v=0,double _c=0):v(_v),c(_c){}
        bool operator <(const qnode &r) const
        {
            return c>=r.c;
        }
};

struct Edge
{
    int v;
    double cost;
    Edge (int _v=0,double _cost = 0):v(_v),cost(_cost){}
};
vector<Edge>E[MAXN];
bool vis[MAXN];
double dis[MAXN];
void Dijkstra(int n,int start)
{
    memset(vis,false,sizeof(vis));
    for(int i = 1; i <= n;i++) dis[i] = INF;
    priority_queue<qnode>que;
    while(!que.empty()) que.pop();
    dis[start] = 0;
    que.push(qnode(start,0));
    qnode tmp;
    while(!que.empty())
    {
        tmp = que.top();
        que.pop();
        int u = tmp.v;
        if(vis[u])continue;
            vis[u] = true;
            for(int i = 0    ;i<E[u].size();i++)
            {
                int v = E[tmp.v][i].v;
                double cost = E[u][i].cost;
                if(!vis[v] && dis[v]>(dis[u]+cost)+wc)
                    {
                        dis[v] = dis[u] + cost;
                        que.push(qnode(v,dis[v]));
                    }
            }
    }
}

void addedge(int u,int v,double w)
{
    E[u].push_back(Edge(v,w));
}
int main()
{
        int n;
        while(cin>>n)
        {
                //memset(dis,0,sizeof(dis));
                memset(E,0,sizeof(E));
                //memset(vis,0,sizeof(vis));
                memset(t,0,sizeof(t));
                memset(L,0,sizeof(L));
                if(n==-1)break;
               // if(!n) {puts("10");continue;}
                double aa,bb,cc,dd,ee;
                int j = 0;
                int ans=1;
                t[ans].x=0;
                t[ans++].y=5;
                for(int i = 1;i <= n;i ++)
                {
                        scanf("%lf%lf%lf%lf%lf",&aa,&bb,&cc,&dd,&ee);
                        L[j].e.x=aa; L[j].e.y=0;L[j].s.x=aa; L[j].s.y=bb;
                        t[ans]=L[j].s;ans++;j++;
                        L[j].e.x=aa; L[j].e.y=cc;
                        t[ans]=L[j].e;ans++;
                        L[j].s.x=aa; L[j].s.y=dd;
                        t[ans++]=L[j].s;j++;
                        L[j].e.x=aa; L[j].e.y=ee;
                        t[ans++]=L[j].e;
                        L[j].s.x=aa; L[j++].s.y=10;

                }
                t[ans].x=10;t[ans].y=5;
                POINT a ;POINT b ;
                //4*n+2个点
                int len = j;
                for(int i = 1;i <= ans; i++)//枚举每一个点
                {
                        for(int  jj = i+1;jj <= ans;jj++)
                        {
                                if(jj!=i)
                                {
                                        LINESEG l;
                                        l.e = t[i];l.s=t[jj];
                                        int flag = 1;
                                        for(int k = 0;k<len;k++)
                                        {
                                                /*POINT aaa = L[k].e;
                                                POINT bbb = L[k].s;
                                                if(dist(aaa,t[i])==0||dist(aaa,t[jj])==0||dist(bbb,t[i])==0||dist(bbb,t[jj])==0)
                                                        ;
                                                else*/
                                                if(intersect_A(L[k],l))
                                                {
                                                        if(i==1) {//cout<<jj<<" "<<k<<endl;
                                                       // cout<<L[k].e.x<<" "<<L[k].e.y<<endl;
                                                       // cout<<L[k].s.x<<" "<<L[k].s.y<<endl;
                                                        }
                                                        flag = 0;break;
                                                }
                                        }
                                        double w = dist(t[i],t[jj]);
                                        if(flag) {addedge(i,jj,w); //cout<<i<<" "<<jj<<endl;
                                        }
                                }
                        }
                }
                Dijkstra(ans,1);
                printf("%.2f\n",dis[ans]);
        }
}