Frogger(图论,最短路径)

来源:https://vjudge.net/problem/POJ-2253

一开始还以为是最短路径,后来看了大佬的博客才发现用的Dijkstra的思想维护最小的最大跳跃距离数组d[],贴一下大佬的博客https://blog.csdn.net/duan_1998/article/details/73716128

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
double w[205][205];
double d[205];
bool vis[205];
int n;
struct node
{
    double x,y;
} no[205];
void djs()
{
    for(int i=1;i<=n;i++)
    {
        d[i]=w[1][i];
        vis[i]=0;
    }
    for(int i=1; i<=n; i++)
    {
        double m=INF;
        int x=-1;
        for(int j=1; j<=n; j++)
            if(!vis[j]&&d[j]<m)
                m=d[x=j];
        if(x!=-1)
        {
            vis[x]=1;//标记
            for(int j=1; j<=n; j++)
                if(!vis[j]&&d[j]>max(d[x],w[x][j]))
                    d[j]=max(d[x],w[x][j]);//更新（维护）d数组
        }
    }
}
int main()
{
    int t=1;
    while(~scanf("%d",&n)&&n)
    {
        for(int i=1; i<=n; i++)
            scanf("%lf%lf",&no[i].x,&no[i].y);
        for(int i=1; i<=n; i++)
            for(int j=i; j<=n; j++)
                w[i][j]=w[j][i]=sqrt((no[i].x-no[j].x)*(no[i].x-no[j].x)+(no[i].y-no[j].y)*(no[i].y-no[j].y));
        djs();
        printf("Scenario #%d\nFrog Distance = %.3lf\n\n",t++,d[2]);//注意格式哦

    }
}