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POJ 3270 Cow Sorting(置换群)
题目链接题意 : N头牛,每个牛的坏脾气都有一个值,每个值都不相同,把这个值按照从小到大排序,如果两个值交换,那么会花掉这两个值之和的时间,让你花最少的时间将每个值从小到大排好序,求最小的总时间。思路 : 这个在黑书上有写,就是置换群,248页有写。写的挺详细的。每个状态都可以分为若干个循环的乘积。...
分类:其他好文   时间:2014-06-09 13:41:20    阅读次数:253
POJ 3615 Cow Hurdles.
Description Farmer John wants the cows to prepare for the county jumping competition, so Bessie and the gang are practicing jumping over hurdles. They are getting tired, though, so they want to be ...
分类:其他好文   时间:2014-06-08 15:55:10    阅读次数:199
POJ 3278 Catch That Cow
Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 44070   Accepted: 13764 Description Farmer John has been informed of the location of a fugitiv...
分类:其他好文   时间:2014-06-08 15:44:43    阅读次数:238
poj 3268 Silver Cow Party , spfa , dijkstra
点击打开链接 两次求最短路(第二次把边反向求) 1、spfa //poj 3268 Silver Cow Party //SPFA #include #include #include using namespace std; const int M = 100000 + 100; const int N = 1000 + 100; const int inf = 1...
分类:其他好文   时间:2014-06-08 05:29:39    阅读次数:241
POJ3278——Catch That Cow
Catch That CowDescriptionFarmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0...
分类:其他好文   时间:2014-06-05 19:54:29    阅读次数:298
poj 3278
Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 43851   Accepted: 13674 Description Farmer John has been informed of the location of a fugitiv...
分类:其他好文   时间:2014-06-05 07:45:38    阅读次数:220
POJ 3268 Silver Cow Party dijkstra单源最短路
裸dijkstra 思路:以x为源点,求到其他点的最短路,之后把邻接矩阵转置,再求一次x源 点的最短路,这样就一次是来的,一次是走的,相加迭代最大值即可 代码: /* poj 3268 8108K 47MS */ #include #include #define MAXN 1005 #define MAX_INT 2147483647 using namespace s...
分类:其他好文   时间:2014-06-04 23:32:57    阅读次数:303
poj 3270 Cow Sorting
http://poj.org/problem?id=3270这道题就是给你一个无序序列转换成有序序列需要花费的代价最小,交换a和b代价为a+b; 1 #include 2 #include 3 #include 4 #define maxn 20000 5 using namespace st...
分类:其他好文   时间:2014-06-03 13:14:32    阅读次数:272
HDU2717-Catch That Cow
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,0...
分类:其他好文   时间:2014-06-03 05:30:23    阅读次数:394
poj3278(Catch That Cow)
题目大意: 一个农主寻找牛。给出农主的位置n和牛的位置k。农主可以通过n-1或者n+1或者n*2的步伐找牛,问至少多少步才能找到自己的牛。解题思路: 简单的BFS。把农主的每一种可能的步伐通过BFS存到栈中,然后看最少多少步到达K坐标。代码: 1 #include 2 #include 3 ...
分类:其他好文   时间:2014-05-31 15:27:27    阅读次数:252
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