Froggy Ford Gym - 100851F

​​http://codeforces.com/gym/100851/attachments​​

青蛙过河 n个石头 当作n个点 任意两点之间都有边 边的花费即为两点欧式距离 现在可以在任意位置加一个石头 使过河的最大边花费最小

先从两岸分别跑一遍最短路 然后n^2的暴力枚举 看新加的石头加在原有石头i和j的中点后的花费是多少即可

注意下一边界即可

#include <bits/stdc++.h>
using namespace std;
const double N=1000000000000000000.0;
const int maxn=1e3+10;

struct node
{
    bool friend operator < (node n1,node n2){
        return n1.val>n2.val;
    }
    int id;
    double val;
};

priority_queue <node> que;
double e[maxn][maxn];
double x[maxn],y[maxn],dis1[maxn],dis2[maxn];
double w;
int book[maxn];
int n;

double getdis(double x1,double y1,double x2,double y2)
{
    return sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
}

void dijkstra(double *dis,int s)
{
    node cur,tmp;
    double w;
    int i,u,v;
    while(!que.empty()) que.pop();
    for(i=1;i<=n;i++) dis[i]=N;
    memset(book,0,sizeof(book));
    dis[s]=0.0;
    tmp.id=s,tmp.val=dis[s];
    que.push(tmp);
    while(!que.empty()){
        cur=que.top();
        que.pop();
        u=cur.id;
        if(book[u]) continue;
        book[u]=1;
        for(v=1;v<=n;v++){
            w=e[u][v];
            if(u!=v&&!book[v]&&dis[v]>max(dis[u],w)){
                dis[v]=max(dis[u],w);
                tmp.id=v,tmp.val=dis[v];
                que.push(tmp);
            }
        }
    }
}

int main()
{
    double ans,px,py;
    int i,j;
    freopen("froggy.in","r",stdin);
    freopen("froggy.out","w",stdout);
    scanf("%lf%d",&w,&n);
    if(n==0){
        printf("%.3f %.3f\n",w/2.0,0.0);
    }
    else{
        for(i=2;i<=n+1;i++) scanf("%lf%lf",&x[i],&y[i]);
        n+=2;
        e[1][n]=e[n][1]=w;
        for(i=2;i<=n-1;i++){
            e[1][i]=e[i][1]=x[i];
            e[i][n]=e[n][i]=w-x[i];
        }
        for(i=2;i<=n-1;i++){
            for(j=i+1;j<=n-1;j++){
                e[i][j]=e[j][i]=getdis(x[i],y[i],x[j],y[j]);
            }
        }
        dijkstra(dis1,1);
        dijkstra(dis2,n);
        ans=N;
        for(i=2;i<=n-1;i++){
            for(j=2;j<=n-1;j++){
                if(i!=j){
                    if(ans>max(e[i][j]/2.0,max(dis1[i],dis2[j]))){
                        ans=max(e[i][j]/2.0,max(dis1[i],dis2[j]));
                        px=(x[i]+x[j])/2.0,py=(y[i]+y[j])/2.0;
                    }
                }
            }
        }
        for(i=2;i<=n-1;i++){
            if(ans>max(x[i]/2.0,dis2[i])){
                ans=max(x[i]/2.0,dis2[i]);
                px=x[i]/2.0,py=y[i];
            }
            if(ans>max(dis1[i],(w-x[i])/2.0)){
                ans=max(dis1[i],(w-x[i])/2.0);
                px=x[i]+(w-x[i])/2.0,py=y[i];
            }
        }
        printf("%.3f %.3f\n",px,py);
    }
    return 0;
}

/*
10 0

10 2
2 1
2 4
*/