Frogger DescriptionFreddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists' sunscreen, he wants to avoid swimm…

题目链接 http://poj.org/problem?id=2253 题意 给出青蛙A,B和若干石头的坐标,现在青蛙A要跳到青蛙B所在的石头上,求出所有路径中最远那一跳的最小值. 思路 Floyd算法的变形,将求两点之间的最短路改成求两点之间最大边权的最小值即可. 代码 #include <algorithm> #include <iostream> #include <cstring> #include <cstdio> #include <cma…

题目链接:https://vjudge.net/problem/POJ-2253 题意:给出n个点的坐标,求点1到点2的forg distance,其定义为点1到点2的所有路径中最长边的最小值. 思路:floyd真的很强大,改一下定义,dis[i][j]表示i到j的frog distance,然后枚举中间点k,转移方程是dis[i][j]=min(dis[i][j],max(dis[i][k],dis[k][j])).复杂度O(n^3). AC代码: e<cstdio> #include<…

题目链接:http://poj.org/problem?id=2253 就是求所有路径的最大边权值的最小值 处理时每次找出距离当前的已选的节点的最短距离,然后更新每个未选节点的值 代码: #include<iostream> #include<cstdlib> #include<cstring> #include<cstdio> #include<cmath> using namespace std; #define maxn 210 #defi…

Description Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists' sunscreen, he wants to avoid swimming and…

UVA - 10048 Audiophobia Consider yourself lucky! Consider yourself lucky to be still breathing and having fun participating in this contest. But we apprehend that many of your descendants may not have this luxury. For, as you know, we are the dweller…

Frogger Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2253 Description Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone…

题目链接:http://poj.org/problem?id=2253 Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 49409   Accepted: 15729 Description Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on a…

做到了这个题,感觉网上的博客是真的水,只有kuangbin大神一句话就点醒了我,所以我写这篇博客是为了让最短路的入门者尽快脱坑...... 本题思路:本题是最短路的变形,要求出最短路中的最大跳跃距离,基本思路与最短路一样,dist数组为当前点到源结点最短路的最大距离,这样的话我们知道只需要改变松弛方程就可以了,每次我们选取一个最小值dist[ k ],那么接下来我们就需要将与结点k相邻的结点都更新,如何更新呢,当然是选取之前所走路中的最大值和现在需要走的路中的值的最大值啦即dist[ j ] =…

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<ctype.h>#include<queue>#include<stack>#include<stdlib.h>#include<math.h>#include<limits.h>#define max(a, b) a>b…