AC日记——Car的旅行路线 洛谷 P1027

Car的旅行路线

思路：

　　这题不难，就是有点恶心；

　　而且，请认真读题目（就是题目卡死劳资）；

来，上代码：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define maxn 807
#define maxm 860005
#define INF 0x7fffffff

struct CityType {
    double x[4],y[4],c;
};
struct CityType city[maxn];

int T,n,s,t,head[maxn],E[maxm],V[maxm];
int cnt,que[maxn<<2];

double W[maxm],dis[maxn],c;

bool if_[maxn];

inline double d(int now,int v1,int v2)
{
    double x1=city[now].x[v1],x2=city[now].x[v2];
    double y1=city[now].y[v1],y2=city[now].y[v2];
    return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
}

inline void edge_add(int u,int v,double w)
{
    E[++cnt]=head[u],V[cnt]=v,W[cnt]=w,head[u]=cnt;
    E[++cnt]=head[v],V[cnt]=u,W[cnt]=w,head[v]=cnt;
}

double ds(int u,int v1,int v,int v2)
{
    double xx=city[u].x[v1]-city[v].x[v2];
    double yy=city[u].y[v1]-city[v].y[v2];
    return sqrt(xx*xx+yy*yy);
}

void spfa()
{
    memset(if_,false,sizeof(if_));
    memset(dis,0x7f,sizeof(dis));
    int h=0,tail=4;que[0]=s*4-3,que[1]=s*4-2;
    que[2]=s*4-1,que[3]=s*4,if_[s*4-3]=true;
    if_[s*4-2]=true,if_[s*4-1]=true,if_[s*4]=true;
    dis[s*4-3]=0,dis[s*4-2]=0,dis[s*4-1]=0,dis[s*4]=0;
    while(h<tail)
    {
        int now=que[h++];if_[now]=false;
        for(int i=head[now];i;i=E[i])
        {
            if(dis[V[i]]>dis[now]+W[i])
            {
                dis[V[i]]=dis[now]+W[i];
                if(!if_[V[i]]) if_[V[i]]=true,que[tail++]=V[i];
            }
        }
    }
}

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        memset(head,0,sizeof(head)),cnt=0;
        scanf("%d%lf%d%d",&n,&c,&s,&t);
        for(int i=1;i<=n;i++)
        {
            for(int v=0;v<3;v++) scanf("%lf%lf",&city[i].x[v],&city[i].y[v]);
            scanf("%lf",&city[i].c);
            double d01=d(i,0,1),d02=d(i,0,2),d12=d(i,1,2);
            int pos=i*4-3;
            if(d01+d02==d12)
            {
                city[i].x[3]=city[i].x[1]+city[i].x[2]-city[i].x[0];
                city[i].y[3]=city[i].y[1]+city[i].y[2]-city[i].y[0];
            }
            else if(d01+d12==d02)
            {
                city[i].x[3]=city[i].x[0]+city[i].x[2]-city[i].x[1];
                city[i].y[3]=city[i].y[0]+city[i].y[2]-city[i].y[1];
            }
            else
            {
                city[i].x[3]=city[i].x[0]+city[i].x[1]-city[i].x[2];
                city[i].y[3]=city[i].y[0]+city[i].y[1]-city[i].y[2];
            }
            for(int j=0;j<=3;j++)
            {
                for(int v=j+1;v<=3;v++) edge_add(pos+j,pos+v,sqrt(d(i,j,v))*city[i].c);
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                int posi=i*4-3,posj=j*4-3;
                for(int v=0;v<4;v++)
                {
                    for(int z=0;z<4;z++) edge_add(v+posi,z+posj,ds(i,v,j,z)*c);
                }
            }
        }
        spfa();double ans=INF;
        for(int i=0;i<=3;i++) ans=min(ans,dis[i+t*4-3]);
        printf("%.1lf",ans);
    }
    return 0;
}