UVA534 Frogger (dijkstra)

题目链接：https://www.luogu.com.cn/problem/UVA534

 

题目大意：求路径中最长的石头间隔最小是多少

 

方法：Dijkstra改变判断条件，d数组改变含义，d[i]为从起点到i路径中最小的间隔

 

示例代码：

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#include <map>
#include <queue>
#include <cmath>
using namespace std;
#define int long long
#define N 4010
#define P pair<double,int>
#define mp make_pair
struct node {double x,y;} a[210];
double m[210][210],d[210];
int n,v[210];
double dis(int i, int j) {
    return sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y));
}
signed main() {
    int tot = 0;
    while(~scanf("%d",&n) && n) {
        memset(m,0,sizeof(m));
        for(int i = 0; i < 210; i++) d[i] = 3000.0;
        memset(v,0,sizeof(v));
        for(int i = 1; i <= n; i++) 
            scanf("%lf%lf",&a[i].x,&a[i].y);
        for(int i = 1; i <= n; i++) 
            for(int j = i+1; j <= n; j++) 
                m[j][i] = m[i][j] = dis(i,j);
        priority_queue<P> q;
        q.push(mp(0,1)); d[1] = 0;
        while(!q.empty()) {
            int u = q.top().second; q.pop();
            if(v[u]) continue;
            v[u] = 1;
            for(int i = 1; i <= n; i++) {
                if(i == u) continue;
                if(!v[i] && max(d[u],m[u][i]) < d[i]) {
                    d[i] = max(d[u],m[u][i]);
                    q.push(mp(-d[i],i));
                }
            }
        }
        printf("Scenario #%d\n",++tot);
        printf("Frog Distance = %.3f\n\n",d[2]);
    }
    return 0;
}