Frogger POJ - 2253 （dijkstra+relax修改）

本题需要维护的不是最短路径的估计值，而是路径最大边权的最小值估计值

因此只需要修改一下relax即可

#include<iostream>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
using namespace std;
#define inf 0x3f3f3f3f
int n;

struct node{
    int num;
    float x,y;
    float d;
    bool operator < (const node & re) const{
        return re.d<d;
    }
};

node nodes[1006];

bool can[1007];
void dijkstra()
{
    memset(can,0,sizeof(can));
    for(int i=1;i<=n;i++) {
        nodes[i].d=inf,nodes[i].num=i;
        cin>>nodes[i].x>>nodes[i].y;
    }
    nodes[1].d=0;
    priority_queue<node>Q;
    Q.push(nodes[1]);
    while(!Q.empty())
    {
        node t=Q.top(); Q.pop();
        if(can[t.num]) continue;
        can[t.num]=1;
        for(int i=1;i<=n;i++){
            if(i==t.num) continue;
            nodes[i].d=min(nodes[i].d,max(nodes[t.num].d,sqrt((nodes[i].x-nodes[t.num].x)*(nodes[i].x-nodes[t.num].x)+(nodes[i].y-nodes[t.num].y)*(nodes[i].y-nodes[t.num].y) ) ));
            Q.push(nodes[i]);
        }
    }
}

int main()
{
    int op=1;
    while(1)
    {
        cin>>n;
        if(n==0) break;
        dijkstra();
        cout<<"Scenario #"<<op<<endl;
        cout<<"Frog Distance = ";
        printf("%.3f\n\n",nodes[2].d);
        op++;
    }
    return 0;
}

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