POJ1556（最短路+线段相交）

题意:给你一些挡板，问从(0,5)->(10,5)的最短路。

题解：暴力枚举所有给的点来判断线段相交，（暴力枚举判），不相交就给建一条边，最后跑一遍最短路

然后吧我的dij不知道为啥跑不过去？或许姿势不对吗..最后用了floyd

#include <iostream>
#include <algorithm>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <set>
#include <cmath>
#define LL long long
#define INF 0x3f3f3f3f
#define MOD 1000000007
const int maxn = 500 + 5;
const double eps = 1e-8;
using namespace std;
bool vis[maxn];
double d[maxn];
double maps[maxn][maxn];
int dcmp(double x){
if(fabs(x) < eps) return 0;
if(x < 0) return -1;
else return 1;
}
struct Point{
double x,y;
Point(){}
Point(double xx, double yy):x(xx),y(yy){}
}pt[maxn][6];
struct Line{
Point s,e;
Line(){}
Line(Point aa, Point bb):s(aa),e(bb){}
};
double cross(Point a, Point b, Point c){
return  (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);
}
double Line_Inter(Line l1, Line l2){//跨立实验
return dcmp(cross(l2.s,l1.s,l1.e)) * dcmp(cross(l2.e,l1.s,l1.e)) < 0;//线段相交
}
double getdis(Point p1, Point p2){
double x1 = p1.x, x2 = p2.x;
double y1 = p1.y, y2 = p2.y;
return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));
}
bool check(Point p1, Point p2, int from, int to){
if(from > to)
return true;
Line l1 = Line(p1,p2);
for(int i=from; i<=to; i++){
Line l2 = Line(pt[i][0],pt[i][1]);
if(Line_Inter(l1, l2))
return false;
l2 = Line(pt[i][2],pt[i][3]);
if(Line_Inter(l1, l2))
return false;
l2 = Line(pt[i][4],pt[i][5]);
if(Line_Inter(l1, l2))
return false;
}
return true;
}
int main(){
int n;
while(scanf("%d",&n) == 1 && n != -1){
int ptcnt = 4*n+1;
for(int i=0; i<=ptcnt; i++)
for(int j=0; j<=ptcnt; j++){
if(i == j) maps[i][j] = maps[j][i] = 0;
else maps[i][j] = maps[j][i] = INF;
}
Point begin = Point(0,5);
Point end = Point(10,5);
for(int i=1; i<=n; i++){
double x,a,b,c,d;
scanf("%lf%lf%lf%lf%lf",&x,&a,&b,&c,&d);
pt[i][0] = Point(x,0);
pt[i][1] = Point(x,a);
pt[i][2] = Point(x,b);
pt[i][3] = Point(x,c);
pt[i][4] = Point(x,d);
pt[i][5] = Point(x,10);
}
if(check(begin,end,1,n)){
printf("10.00\n");
continue;
}
for(int i=1; i<=n; i++){
for(int k=1; k<=4; k++){
if(check(begin,pt[i][k],1,i-1)){
double c = getdis(begin,pt[i][k]);
maps[0][(i-1)*4+k] = maps[(i-1)*4+k][0] = c;
}
if(check(end,pt[i][k],i+1,n)){
double c = getdis(end,pt[i][k]);
maps[4*n+1][(i-1)*4+k] = maps[(i-1)*4+k][4*n+1] = c;
}
}
}
for(int i=1; i<n; i++)
for(int k=i+1; k<=n; k++){
for(int p=1; p<=4; p++){
for(int q=1; q<=4; q++){
if(check(pt[i][p],pt[k][q],i+1,k-1)){
double c = getdis(pt[i][p],pt[k][q]);
maps[(i-1)*4+p][(k-1)*4+q] = maps[(k-1)*4+q][(i-1)*4+p] = c;
}
}
}
}
for(int k=0; k<=ptcnt; k++)
for(int i=0; i<=ptcnt; i++)
for(int j=0; j<=ptcnt; j++)
if(maps[i][k] + maps[k][j] < maps[i][j])
maps[i][j] = maps[i][k] + maps[k][j];
printf("%.2f\n",maps[0][ptcnt]);
}
}