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…

题意:要从起点的石头跳到终点的石头,设The frog distance为从起点到终点的某一路径中两点间距离的最大值,问在从起点到终点的所有路径中The frog distance的最小值为多少. 分析: 解法一:Dijkstra,修改最短路模板,d[u]表示从起点到u的所有路径中两点间距离的最大值的最小值. #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #incl…

Time Limit: 1000MS Memory Limit: 65536K 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' s…

链接: http://poj.org/problem?id=2253 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#problem/D Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 21206   Accepted: 6903 Description Freddy Frog is sitting on a stone in the…

点击打开链接 Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 21653   Accepted: 7042 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 visi…

题目 这里的dijsktra的变种代码是我看着自己打的,终于把代码和做法思路联系上了,也就是理解了算法——看来手跟着画一遍真的有助于理解. #define _CRT_SECURE_NO_WARNINGS #include<string.h> #include<stdio.h> #include<math.h> #include<algorithm> using namespace std; ; #define typec double const typec…

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 instead rea…

题意:想给你公青蛙位置,再给你母青蛙位置,然后给你剩余位置,问你怎么走,公青蛙全力跳的的最远距离最小. 思路:这里不是求最短路径,而是要你找一条路,青蛙走这条路时,对他跳远要求最低.这个思想还是挺好迁移的,原来我们用mp[i][j]表示i到j最短路径,那么我们现在用它表示i到j最大步伐,然后每次比较,只要最大步伐比他小,那么我们就走新的路.注意最后是mp[1][2],一直mp[1][n]没改无限WA. 代码; #include<cstdio> #include<set> #incl…

POJ 2253 Frogger 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 swimmin…

POJ. 2253 Frogger (Dijkstra ) 题意分析 首先给出n个点的坐标,其中第一个点的坐标为青蛙1的坐标,第二个点的坐标为青蛙2的坐标.给出的n个点,两两双向互通,求出由1到2可行通路的所有步骤当中,步长最大值. 在dij原算法的基础上稍作改动即可.dij求解的是单源最短路,现在求解的是步长最大值,那么更新原则就是,当前的这一步比保存的步如果要大的话,就更新,否则就不更新. 如此求解出来的就是单源最大步骤. 代码总览 #include <cstdio> #include &…