Fiber Network Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2725   Accepted: 1252 Description Several startup companies have decided to build a better Internet, called the "FiberNet". They have already installed many nodes that ac…

题目链接:http://poj.org/problem?id=2263 题意:题中给出相连通不同城市之间的载货量,要求找到一条从指定起点到终点的路径,并满足载货量最大. #include <iostream> #include <cstdio> #include <cstring> #include <string> using namespace std; ; int dis[POINT][POINT]; string city[POINT]; int m…

Description Big Johnsson Trucks Inc. is a company specialized in manufacturing big trucks. Their latest model, the Godzilla V12, is so big that the amount of cargo you can transport with it is never limited by the truck itself. It is only limited by…

Heavy Cargo Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3768   Accepted: 2013 Description Big Johnsson Trucks Inc. is a company specialized in manufacturing big trucks. Their latest model, the Godzilla V12, is so big that the amount…

E - Arbitrage Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2240 Appoint description:  System Crawler  (2015-11-24) Description Arbitrage is the use of discrepancies in currency exchange rates…

题目链接:http://poj.org/problem?id=1125 多源点最短路中的,最长路的,最短路. 看到这里就懵逼了,解释一下,找到一个源点,使得路最短,(遍历源点),路最短怎么求呢? 就是找到从该源点出发,到达所有点中的最长的点的路径,就是他的最短路,然后根据n个源点,找到这样的最长路最短的,就行了. 这里的具体Floyd算法,我还没推过,dis[i][j]从I到J的最短路. #include <stdio.h> #include <string.h> #include…

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…

题目链接:http://poj.org/problem?id=3615 题意:大致题意:有N个木桩,和M个木桩对之间的高度差(从x跳到y需要往上跳的高度).从x跳跃到y的路径消耗的体力值是路径中的一个最大高度差.求一条消耗体力最小的路径. #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; #define INF 0x3f3f3f3f ][]; int main…

题目:click here 题意: 给出两只青蛙的坐标A.B,和其他的n-2个坐标,任意两坐标间是双向连通的.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. 分析: 最小化最大值-------- #include <iostream> #include <cstdio> #include <cstring> #…

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors. The contest is conduct…