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…

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…

题意:有两只青蛙,a在第一个石头,b在第二个石头,a要到b那里去,每种a到b的路径中都有最大边,求所有这些最大边的最小值.思路:将所有边长存起来,排好序后,二分枚举答案. 时间复杂度比较高,344ms. #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <math.h> using namespace std; ; co…

题意:给定n个点的坐标,问从第一个点到第二个点的最小跳跃范围.d(i)表示从第一个点到达第i个点的最小跳跃范围. AC代码 #include <cstdio> #include <cmath> #include <cctype> #include <algorithm> #include <cstring> #include <utility> #include <string> #include <iostream…

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 &…

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…

Til the Cows Come Home 题目链接: http://acm.hust.edu.cn/vjudge/contest/66569#problem/A Description The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2<=n<=200). The next n lines each con…

嗯....   dijkstra是求最短路的一种算法(废话,思维含量较低,   并且时间复杂度较为稳定,为O(n^2),   但是注意:!!!!         不能处理边权为负的情况(但SPFA可以处理,今后会讲)   借一个何大佬的图,因为会在代码中提到红.绿.空三种颜色,以及小v,   通过图会比较清晰一些: 思路大约明白了下面就呈上带批注模板代码: #include <cstdio>//dijkstra求最短路 #include <cstring> #include <…