poj 1556 The Doors （bellman+几何）

The Doors

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 6294 Accepted: 2530

Description

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 file 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

15 4 6 7 824 2 7 8 97 3 4.5 6 7-1

Sample Output

10.0010.06

有些点中间有墙不能连线，故要挨个判断，然后建图，用最短路解决。

</pre><pre name="code" class="cpp">#include"stdio.h"#include"string.h"#include"math.h"#include"iostream"#include"algorithm"using namespace std;#define LL __int64#define N 100#define M 1000005const double inf=1e10;#define max(a,b) (a>b?a:b)struct point  //存储点坐标{    double x,y;}p[N];struct edge   //存储边信息{    int u,v;}e[N*N];int i,j,n,psize,esize;double wx[20];  //存储每堵墙的X坐标double g[N][N]; //邻接矩阵double py[20][4];double d[N];double Dis(point a,point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}//判断点（x3,y3）是否位于点(x1,y1)和(x2,y2)所确定直线的上方还是下方  //返回值>0表示 (x3,y3)位于直线上方(逆时针方向)，<0表示位于下方 （顺时针方向） double Cross(double x1,double y1,double x2,double y2,double x3,double y3){    return (x2-x1)*(y3-y1)-(x3-x1)*(y2-y1);}bool Isok(point a,point b){                 if(a.x>=b.x)        return false;    int i=0;    while(wx[i]<=a.x&&i<n)        i++;    while(wx[i]<b.x&&i<n)    {        if(Cross(a.x,a.y,b.x,b.y,wx[i],0)*Cross(a.x,a.y,b.x,b.y,wx[i],py[i][0])<0           ||Cross(a.x,a.y,b.x,b.y,wx[i],py[i][1])*Cross(a.x,a.y,b.x,b.y,wx[i],py[i][2])<0           ||Cross(a.x,a.y,b.x,b.y,wx[i],py[i][3])*Cross(a.x,a.y,b.x,b.y,wx[i],10)<0)        {            return false;        }        i++;    }    return true;}void Bellman(){    int i,j;    for(i=0;i<psize;i++)        d[i]=inf;    d[0]=0;    bool ff=true;    int u,v;    for(i=0;i<psize&&ff;i++)    {        ff=false;        for(j=0;j<esize;j++)        {            u=e[j].u; v=e[j].v;            if(d[u]<inf&&d[u]+g[u][v]<d[v])            {                d[v]=d[u]+g[u][v];                ff=true;            }        }    }    printf("%.2f\n",d[psize-1]);}void solve(){    int i,j;    p[0].x=0;    p[0].y=5;    psize=1;    for(i=0;i<n;i++)    {        scanf("%lf",&wx[i]);        for(j=0;j<4;j++)        {            p[psize].x=wx[i];            scanf("%lf",&py[i][j]);            p[psize++].y=py[i][j];        }    }    p[psize].x=10;    p[psize++].y=5;    for(i=0;i<psize;i++)    {        for(j=0;j<psize;j++)            g[i][j]=inf;    }    esize=0;    for(i=0;i<psize;i++)    {        for(j=i+1;j<psize;j++)        {            if(Isok(p[i],p[j]))            {                g[i][j]=Dis(p[i],p[j]);                e[esize].u=i;                e[esize++].v=j;            }        }    }    Bellman();}int main(){    while(scanf("%d",&n)!=-1)    {        if(n==-1)            break;        solve();    }    return 0;}