Til the Cows Come Home Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 38556   Accepted: 13104 Description Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the…

最短路 Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 45288    Accepted Submission(s): 19993 Problem Description 在每年的校赛里,所有进入决赛的同学都会获得一件很漂亮的t-shirt.但是每当我们的工作人员把上百件的衣服从商店运回到赛场的时候,却是非常累的!所以现在他们想要寻找…

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible. Farmer John's field ha…

题意:给出一堆双向路,求从N点到1点的最短路径,最裸的最短路径,建完边之后直接跑dij或者spfa就行 dij: #include<stdio.h> #include<string.h> #include<queue> #include<algorithm> #include<vector> using namespace std; typedef pair<int,int> pii; const int INF=0x3f3f3f3f…

MPI Maelstrom Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7471   Accepted: 4550 Description BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchic…

hud1548 a strange lift  最短路/bfs  题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1548 题意:一个奇怪的电梯,每层楼的按键 只能上达 i + k[i] 层,下至 i- k[i] 层.不能到达超过n楼, 也不能小于 1楼.问最少按键数.以单个0结束输入. 思路:bfs, 从起点出发,每个楼层只会访问一次,在不出界的情况下访问能够到达的楼层. 总结:布吉岛神马原因,只有用G++交才过,导致我纠结了好多天,[摔]! AC…

Dijkstra+ 链式前向星+ 优先队列   Dijkstra算法 Dijkstra最短路算法,个人理解其本质就是一种广度优先搜索.先将所有点的最短距离Dis[ ]都刷新成∞(涂成黑色),然后从起点x (Dis[x]= 0, Dis[]值最小 )开始查询:先将x 加入(涂成灰色),对x 的所有边进行遍历,对所有搜索到的点x+ 1 进行松弛(刷新),若经过x 点的松弛,得到的距离小于原来的值:Dis[x]+ dis(x, x+ 1) < Dis[x+ 1], 则用新值刷新,把x+ 1加入(涂成灰…

AC代码 POJ2387 Til the Cows Come Home Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly…

Til the Cows Come Home POJ-2387 这题是最简单的最短路求解题,主要就是使用dijkstra算法,时间复杂度是\(O(n^2)\). 需要注意的是,一定要看清楚题目的输入要求,是先输入边,再输入顶点,一开始我没看清,wrong answer了一次. package POJ; import java.util.*; public class POJ_2387 { private static int n,t; static int [][]w; static boole…

前两天自学了一点点最短路..看起来很简单的样子... 就去kuangbin的专题找了最简单的一道题练手..然后被自己萌萌的三重for循环超时虐的不要不要的~ 松弛虽然会但是用的十分之不熟练... 代码~ #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int tance[1005][1005]; bool vis[1005]; int dis[1005]; int…