POJ 1556 The Doors

计算几何+最短路

枚举线段是否相交建图，然后跑最短路

#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;

const int maxn=1000+10;
const double eps=1e-8;
int n;
int totP,totL;

struct point
{
    double x;
    double y;
} p[1000];
struct Line
{
    point a;
    point b;
} line[1000];
struct Path
{
    int to;
    double val;
    Path(int t,double v)
    {
        to=t;
        val=v;
    }
};
vector<Path>G[maxn];
queue<int>Q;
bool flag[maxn];
double d[maxn];

double xmult(point p1,point p2,point p0)
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}

int opposite_side(point p1,point p2,point l1,point l2)
{
    return xmult(l1,p1,l2)*xmult(l1,p2,l2)<-eps;
}

int intersect_ex(point u1,point u2,point v1,point v2)
{
    return opposite_side(u1,u2,v1,v2)&&opposite_side(v1,v2,u1,u2);
}

void read()
{
    totP=0,totL=0;
    p[totP].x=0,p[totP++].y=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);
        p[totP+0].x=x,p[totP+0].y=0;
        p[totP+1].x=x,p[totP+1].y=a;
        p[totP+2].x=x,p[totP+2].y=b;
        p[totP+3].x=x,p[totP+3].y=c;
        p[totP+4].x=x,p[totP+4].y=d;
        p[totP+5].x=x,p[totP+5].y=10;

        line[totL].a=p[totP+0];
        line[totL].b=p[totP+1];
        totL++;
        line[totL].a=p[totP+2];
        line[totL].b=p[totP+3];
        totL++;
        line[totL].a=p[totP+4];
        line[totL].b=p[totP+5];
        totL++;

        totP=totP+6;
    }
    p[totP].x=10,p[totP++].y=5;
}

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 SPFA()
{
    memset(flag,0,sizeof flag);
    for(int i=0; i<=1000; i++) d[i]=9999999;
    while(!Q.empty()) Q.pop();
    Q.push(0);
    flag[0]=1;
    d[0]=0;
    while(!Q.empty())
    {
        int head=Q.front();
        Q.pop();
        for(int i=0; i<G[head].size(); i++)
        {
            Path path=G[head][i];
            if(d[head]+path.val<d[path.to])
            {
                d[path.to]=d[head]+path.val;
                if(!flag[path.to])
                {
                    flag[path.to]=1;
                    Q.push(path.to);
                }
            }
        }
    }
}

int main()
{
    while(~scanf("%d",&n))
    {
        if(n==-1) break;
        read();

        for(int i=0; i<=1000; i++) G[i].clear();

        for(int i=0; i<totP; i++)
        {
            for(int j=i+1; j<totP; j++)
            {
                if(abs(p[i].x-p[j].x)<eps) continue;
                bool fail=0;
                for(int k=0; k<totL; k++)
                    if(intersect_ex(p[i],p[j],line[k].a,line[k].b))
                        fail=1;
                if(!fail)
                {
                    double len=dis(p[i],p[j]);
                    Path path(j,len);
                    G[i].push_back(path);
                }
            }
        }
        SPFA();

        printf("%.2f\n",d[totP-1]);
    }
    return 0;
}