POJ 1556 (toj 1952)The Doors

The Doors

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Time Limit: 1.0 Seconds   Memory Limit: 65536K
Total Runs: 126   Accepted Runs: 69

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The Problem

You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x = 10, y = 0, and y = 10. The initial and final points of the path are always (0, 5) and (10, 5). There will also be from 0 to 18 vertical walls inside the chamber, each with two doorways. The figure below illustrates such a chamber and also shows the path of minimal length.

 

 

Input

The input data for the illustrated chamber would appear as follows.

2
4 2 7 8 9
7 3 4.5 6 7

The first line contains the number of interior walls. Then there is a line for each such wall, containing five real numbers. The first number is the x coordinate of the wall (0 < x < 10), and the remaining four are the y coordinates of the ends of the doorways in that wall. The x coordinates of the walls are in increasing order, and within each line the y coordinates are in increasing order. The input will contain at least one such set of data. The end of the data comes when the number of walls is -1.

Output

The output should contain one line of output for each chamber. The line should contain the minimal path length rounded to two decimal places past the decimal point, and always showing the two decimal places past the decimal point. The line should contain no blanks.

Sample Input

1
5 4 6 7 8
2
4 2 7 8 9
7 3 4.5 6 7
-1

Sample Output

10.00
10.06

 

 

Source: Mid-Central USA 1996

 

 

由于交题的时候poj挂了，在hdu上没找到，突然想到自己学校的oj。在toj上一找，发现有这道题，就交了一发，1A，很开心。

发现toj还是挺厉害的，题也是很全的~

思路：乍一看是一道几何题，再一想突然发现可以用最短路做

每个点之间连线，如果不和任何墙相交就建边

最短路跑一下就行了

 

#include <iostream>
#include <cstring>
#include <queue>
#include <cmath>
#include <iomanip>

#define INF 100000000.0

using namespace std;

struct point
{
    double x,y;
    point(){};
    point (double _x,double _y)
    {
        x=_x,y=_y;
    }
};
typedef point vec;

double operator /(vec a, vec b)
{
    return (a.x*b.y-a.y*b.x);
}
vec operator- (point a,point b)
{
    vec c;
    c.x=a.x-b.x;
    c.y=a.y-b.y;
    return c;
}
struct line
{
    point a,b;
    line(){};
    line (point _a,point _b)
    {
        a=_a;
        b=_b;
    }
};
double dis(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

struct edge
{
    int v,next;
    double w;
};
int cross(line m,line n)
{
    if(((n.a-m.a)/(m.b-m.a))*((n.b-m.a)/(m.b-m.a))<=0
       &&((m.a-n.a)/(n.b-n.a))*((m.b-n.a)/(n.b-n.a))<=0)
       return 1;
       else return 0;
}
edge e[5005];
int head[105],cnt=0,vis[5005],n,m;
double dist[105];
queue<int>q;
void addedge(int u,int v,double w)
{
    e[cnt].v=v;
    e[cnt].w=w;
    e[cnt].next=head[u];
    head[u]=cnt++;
}
void init()
{
    memset(head,-1,sizeof(head));
    cnt=0;
}
void spfa(int s)
{
    for(int i=0;i<=n;i++)
        dist[i]=INF,vis[i]=0;
    while(!q.empty()) q.pop();
    q.push(s);
    vis[s]=1;
    dist[s]=0.0;
    while(!q.empty())
    {
        int tmp;
        tmp=q.front();
        q.pop();
        for(int i=head[tmp];i!=-1;i=e[i].next)
        {
            if(dist[e[i].v]>dist[tmp]+e[i].w)
            {
                dist[e[i].v]=dist[tmp]+e[i].w;
                if(!vis[e[i].v])
                {
                    vis[e[i].v]=1;
                    q.push(e[i].v);
                }
            }

        }
        vis[tmp]=0;
    }
}
point p[105];
line le[5005];
bool check(int i,int j)
{
    line temp(p[i],p[j]);
    int left,right;
    if(i!=0 && j!=n-1)
    {
        left=(((i-1)/4)+1)*3;
        right=((j-1)/4)*3;
        for(int c1=left;c1<right;c1++)
        {
            if(cross(temp,le[c1]))
            {
                return false;
            }
        }
    }
    else if(i==0 && j!=n-1)
    {
        right=((j-1)/4)*3;
        for(int c1=0;c1<right;c1++)
        {
            if(cross(temp,le[c1]))
            {
                return false;
            }
        }
    }
    else if(i!=0 && j==n-1)
    {
        left=(((i-1)/4)+1)*3;
        for(int c1=left;c1<m-1;c1++)
        {
            if(cross(temp,le[c1]))
            {
                return false;
            }
        }
    }
    else
    {
        for(int c1=0;c1<m-1;c1++)
        {
            if(cross(temp,le[c1]))
            {
                return false;
            }
        }
    }
    return true;
}
int main()
{
    int t;
    while(cin>>t)
    {
        if(t==-1)
            break;
        init();
        int totl=0,totp=1;
        n=4*t+2;
        m=3*t;
        p[0].x=0,p[0].y=5.0;
        p[n-1].x=10.0,p[n-1].y=5.0;
        for(int i=0;i<t;i++)
        {
            double x,y1,y2,y3,y4;
            point a,b;
            line tmp;
            cin>>x>>y1>>y2>>y3>>y4;
            p[totp].x=x,p[totp++].y=y1;
            p[totp].x=x,p[totp++].y=y2;
            p[totp].x=x,p[totp++].y=y3;
            p[totp].x=x,p[totp++].y=y4;
            a.x=x,a.y=0.0;
            b.x=x,b.y=y1;
            tmp.a=a,tmp.b=b;
            le[totl++]=tmp;
            a.x=x,a.y=y2;
            b.x=x,b.y=y3;
            tmp.a=a,tmp.b=b;
            le[totl++]=tmp;
            a.x=x,a.y=y4;
            b.x=x,b.y=10.0;
            tmp.a=a,tmp.b=b;
            le[totl++]=tmp;
        }
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                if(check(i,j))
                    addedge(i,j,dis(p[i],p[j]));
            }
        }
        spfa(0);
        double ans=dist[n-1];
        cout<<fixed<<setprecision(2)<<ans<<endl;
    }
    return 0;
}

这是16年的还是17年写的 csdn竟然没给我发出去 。。。一口老血