Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22557   Accepted: 7339 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,…

题目传送门 /* 最短路:Floyd算法模板题 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include <string> #include <vector> using namespace std; + ; const int INF = 0x3f3f3f3f; do…

POJ 2253 Frogger题目意思就是求所有路径中最大路径中的最小值. #include<iostream> #include<cstdio> #include<string.h> #include <utility>//make_pair的头文件 #include<math.h> using namespace std; ; double map[maxn][maxn]; int n; typedef struct pair<int…

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 &…

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…

题目链接:http://poj.org/problem?id=2253 Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 38366   Accepted: 12357 Description Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on a…

传送门: http://poj.org/problem?id=2253 Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 58328   Accepted: 18293 Description Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on a…

题目链接:http://poj.org/problem?id=2253 Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 25773   Accepted: 8374 Description Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on an…

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…

原题链接:http://poj.org/problem?id=2253 Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 30637   Accepted: 9883 Description Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on an…

链接:poj 2253 题意:给出青蛙A,B和若干石头的坐标,现青蛙A想到青蛙B那,A可通过随意石头到达B, 问从A到B多条路径中的最长边中的最短距离 分析:这题是最短路的变形,曾经求的是路径总长的最小值,而此题是通路中最长边的最小值,每条边的权值能够通过坐标算出,由于是单源起点,直接用SPFA算法或dijkstra算法就能够了 SPFA 16MS #include<cstdio> #include<queue> #include<cmath> #include<…

题目链接: http://poj.org/problem?id=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. He plans to visit her, but since the water is dirty and full of tourists' suns…

Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 24879   Accepted: 8076 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,…

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…

http://poj.org/problem?id=2253 #include <iostream> #include <queue> #include <cmath> #include <iomanip> using namespace std; queue <int > que; int co[202][2]; double d[202][202],u[202][202]; int n; int main(){ ios::sync_with_…

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…

http://poj.org/problem?id=2253 题意 : 题目是说,有这样一只青蛙Freddy,他在一块石头上,他呢注意到青蛙Fiona在另一块石头上,想去拜访,但是两块石头太远了,所以他只有通过别的石头跳过去,所以,从他的石头到Fiona的石头每一条可走的路,假设是n条,就需要你求出frog distance,这个所谓的距离就是指这n条路中,每条路选取组成这条路中最长的那边,最后一共有n条边,找这n条边里最短的那一条输出. 思路 : 就是一个最短路的问题,不过不需要求最短路的权值…

题目:http://poj.org/problem?id=2253 题意:给出两只青蛙的坐标A.B,和其他的n-2个坐标,任一两个坐标点间都是双向连通的.显然从A到B存在至少一条的通路,每一条通路的元素都是这条通路中前后两个点的距离,这些距离中又有一个最大距离. 现在要求求出所有通路的最大距离,并把这些最大距离作比较,把最小的一个最大距离作为青蛙的最小跳远距离. #include <iostream> #include<cstdio> #include<cstring>…

题目链接:http://poj.org/problem?id=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. He plans to visit her, but since the water is dirty and full of tourists' sunsc…

http://poj.org/problem?id=2253 题意: 有两只青蛙A和B,现在青蛙A要跳到青蛙B的石头上,中间有许多石头可以让青蛙A弹跳.给出所有石头的坐标点,求出在所有通路中青蛙需要跳跃距离的最小值. 思路: dijkstra算法的变形.本来是dist是记录最短距离,在这道题中可以把它变为已经跳过的最大距离,稍微改一下松弛算法就可以.具体见代码. #include<iostream> #include<algorithm> #include<string>…

Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 24979   Accepted: 8114 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,…

题目链接:http://poj.org/problem?id=2253 题目意思:找出从Freddy's stone  到  Fiona's stone  最短路中的最长路. 很拗口是吧,举个例子.对于 i 到 j 的一条路径,如果有一个点k, i 到 k 的距离 && k 到 j 的距离都小于 i 到 j 的距离,那么就用这两条中较大的一条来更新 i 到 j 的距离 .每两点之间都这样求出路径.最后输出 1 到 2 的距离(1:Freddy's stone   2:Fiona's sto…

传送门 Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 39453   Accepted: 12691 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…

题意: 题目撰写者的英语真是艰难晦涩,看了别人题解,才知道这题题意. 两个forger 一个froger 要蹦到另外一个froger处,他们的最短距离是这样定义的 : The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between t…

这题的坑点在POJ输出double不能用%.lf而要用%.f...真是神坑. 题意:给出一个无向图,求节点1到2之间的最大边的边权的最小值. 算法:Dijkstra 题目每次选择权值最小的边进行延伸访问,最坏情况下每条路径都要访问,复杂度O(n^2) 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm>…

Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 28333   Accepted: 9208 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,…

Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 30427   Accepted: 9806 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,…

Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions:57696   Accepted: 18104 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,…

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…

题意  给你n个点的坐标  求第1个点到第2个点的全部路径中两点间最大距离的最小值 非常水的floyd咯 #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> using namespace std; const int N=205; double d[N][N]; int x[N],y[N],n; void floyd() { for(int k=1;k<=n;+…