http://poj.org/problem?id=1556
The Doors
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 6120 | Accepted: 2455 |
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
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
开始的时候真的是二逼了，
1、判断相交的函数写错了，我居然判断的是是不是跟源点和终点的直线相交。。。二逼啊，，，
2、然后改了之后还wa,因为判断里少了个！，，，，没取反，，，
3、极限的点，比如每道墙的最上沿和最下沿，这两个点不可达，就是说从源头到终点不能经过这两个点，开始的时候没排除，虽然那样的话也能AC,还是题目数据太弱了啊
我自己写的判断直线相交的模板：
/*==========================================================*|| 判断点在直线上或直线相交
1、函数值为0，表示在直线上；
2、test(a,b,t1)*test(a,b,t2)<0表示直线ab和直线t1t2相交
\*==========================================================*/
double test(Point a,Point b, Point t)
{
return (b.y-a.y)*(t.x-b.x)-(b.x-a.x)*(t.y-b.y);
}
思路还是比较顺的，就是最短路+判断直线相交
贴代码：
#include<cstdio>
#include<cstring>
#include <string>
#include <map>
#include <iostream>
#include <cmath>
using namespace std;
#define INF 10000
const double eps=1e-6;
const int MAXN = 1011;
#define Max(a,b) (a)>(b)?(a):(b)
int cntp;
int wn;
struct Point{
Point(double x=0,double y=0):x(x),y(y){}
double x,y;
int id;
}p[MAXN];
struct Wall{
double s1,e1;
double s2,e2;
double s3;
}w[20];//=0~~=wn
double e[MAXN][MAXN],dist[MAXN];
double dis(Point a, Point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
void init()
{
cntp=1;
for(int i=0;i<=MAXN;i++)
for(int j=0;j<=MAXN;j++)
{
if(i == j)e[i][j]=0;
else e[i][j]=INF;
}
p[0].x=0,p[0].y=5,p[0].id=0;
for(int i=0;i<=MAXN;i++)dist[i]=INF;
}
double test(Point a,Point b, Point t)
{
return (b.y-a.y)*(t.x-b.x)-(b.x-a.x)*(t.y-b.y);
}
bool Judge(Point a, Point b)
{
if(a.id>b.id)
{
Point t=a;
a=b;
b=t;
}
//int flag=1;
if(a.id>0)
if(a.y -0.0 <=eps||10.0-a.y <=eps)
return 0;
if(b.id<cntp-1)
if(b.y-0.0<=eps || 10.0-b.y<=eps)
return 0;
for(int i=a.id+1;i<b.id;i++)
{
Point p1(w[i].s1,w[i].e1),p2(w[i].s1,w[i].s2),p3(w[i].s1,w[i].e2),p4(w[i].s1,w[i].s3);
if(!(
test(a,b,p1)*test(a,b,p2)<0 ||
test(a,b,p3)*test(a,b,p4)<0)
)return 0;
}
/*for(int i=a.id+1;i<b.id;i++)
{
if(!(
(w[i].e1<5.0&&w[i].s2>5.0)
|| (w[i].e2<5.0&&w[i].s3>5.0)
)
)return 0;
}*/
return 1;
}
void Build()
{
for(int i=0;i<cntp;i++)
{
for(int j=i+1;j<cntp;j++)
{
//if(i == j)continue;
if(p[i].id == p[j].id)continue;
if(Judge(p[i],p[j]))
{
e[i][j]=min(e[i][j],dis(p[i],p[j]));
}
}
}
}
void Bellman(int v0)
{
int n=cntp;
for(int i=0;i<cntp;i++)
{
dist[i]=e[v0][i];
//if(i!=v0 && dist[i]<INF)
}
for(int k=2;k<n;k++)
{
for(int u=0;u<n;u++)
{
if(u!=v0)
{
for(int j=0;j<n;j++)
{
if(e[j][u]!=INF && dist[j]+e[j][u]<dist[u])
{
dist[u]=dist[j]+e[j][u];
}
}
}
}
}
}
int main()
{
// freopen("poj1556.txt","r",stdin);
while(~scanf("%d",&wn) && ~wn)
{
init();
for(int i=1;i<=wn;i++)
{
scanf("%lf%lf%lf%lf%lf",&w[i].s1,&w[i].e1,&w[i].s2,&w[i].e2,&w[i].s3);
p[cntp].id=p[cntp+1].id=p[cntp+2].id=p[cntp+3].id=p[cntp+4].id=p[cntp+5].id=i;
p[cntp].x=p[cntp+1].x=p[cntp+2].x=p[cntp+3].x=p[cntp+4].x=p[cntp+5].x=w[i].s1;
p[cntp].y=0.0,p[cntp+1].y=w[i].e1,p[cntp+2].y=w[i].s2,p[cntp+3].y=w[i].e2,p[cntp+4].y=w[i].s3,p[cntp+5].y=10.0;
////////////////
//e[cntp+1][cntp+2]=w[i].s2-w[i].e1;
// e[cntp+3][cntp+4]=w[i].s3-w[i].e2;
cntp+=6;
}
p[cntp].x=10.0,p[cntp].y=5.0,p[cntp].id=++wn;
cntp++;
//if()
Build();
Bellman(0);
printf("%.2lf\n",dist[cntp-1]);
///////////////////
// for(int i=0;i<cntp;i++)
// printf("%d %lf\n",i,dist[i]);
}
return 0;
}
poj 1556 zoj1721 BellmanFord 最短路+判断直线相交