[kuangbin带你飞]专题四 最短路练习
文章平均质量分 78
Strive_Buff
面向未来,才能创造未来。
展开
专栏收录文章
- 默认排序
- 最新发布
- 最早发布
- 最多阅读
- 最少阅读
-
poj 1860 Currency Exchange ([kuangbin带你飞]专题四 最短路练习)
DescriptionSeveral currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these curren原创 2016-11-10 15:40:12 · 570 阅读 · 0 评论 -
poj 3268 Silver Cow Party ([kuangbin带你飞]专题四 最短路练习)
DescriptionOne cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirect原创 2016-11-10 15:49:07 · 464 阅读 · 0 评论 -
POJ 1797 Heavy Transportation
DescriptionBackground Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from t原创 2016-11-10 16:29:01 · 365 阅读 · 0 评论 -
poj 3660 Cow Contest ([kuangbin带你飞]专题四 最短路练习)
DescriptionN (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原创 2016-11-02 15:45:23 · 442 阅读 · 0 评论 -
poj 1502 MPI Maelstrom([kuangbin带你飞]专题四 最短路练习)
DescriptionBIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchical communication subsystem. Valentine McKee'原创 2016-11-01 17:10:40 · 491 阅读 · 0 评论 -
poj 3259 Wormholes ([kuangbin带你飞]专题四 最短路练习)
DescriptionWhile exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination原创 2016-11-01 11:28:23 · 544 阅读 · 0 评论 -
poj 2240 Arbitrage ([kuangbin带你飞]专题四 最短路练习)
DescriptionArbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar原创 2016-11-02 18:33:04 · 836 阅读 · 0 评论 -
poj 3026 Borg Maze
题目大意:s会分裂,然后从s到A的最小的步是多少。解题思路:先bfs算出A或者S到任何点的距离,然后最求最小生成树,用dij或者prim都可以#include#include#include#include#includeusing namespace std;const int maxn = 100+10;char mp[maxn][maxn];int c原创 2016-11-15 10:24:20 · 411 阅读 · 0 评论 -
poj 1679 The Unique MST ([kuangbin带你飞]专题八 生成树 )
题目大意:给你一个无向图,问如果这个无向图的最小生成树是唯一的就输出最小生成树的权值,否则输出Not Unique!解题思路,求次小生成树,如果次小生成树中有何最小的权值一样的,则生成树不唯一,否则唯一用Kruskal,求最小生成树,然后枚举其中一个边去掉以后(n-1条),再求最小生成树,然后和原来的权值比较,#include#include#includeusing nam原创 2016-11-15 20:00:53 · 477 阅读 · 0 评论
分享