Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4867    Accepted Submission(s): 1329 Problem Description Rompire is a robot kingdom and a lot of robots live there peacefully. But one day, the ki…

Problem Description Rompire is a robot kingdom and a lot of robots live there peacefully. But one day, the king of Rompire was captured by human beings. His thinking circuit was changed by human and thus became a tyrant. All those who are against him…

Prison Break Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3214    Accepted Submission(s): 829 Problem Description Rompire is a robot kingdom and a lot of robots live there peacefully. But on…

Problem Description Rompire . Now it’s time to escape, but Micheal# needs an optimal plan and he contacts you, one of his human friends, for help. The jail area is a rectangle contains n×m little grids, each grid might be one of the following: ) Empt…

Prison Break 时间限制: 1 Sec  内存限制: 128 MB提交: 105  解决: 16[提交][状态][讨论版] 题目描述 Scofild又要策划一次越狱行动,和上次一样,他已经掌握了整个监狱的地图,看守的位置,以及准备好了逃出监狱的出口.由于消息被其他监狱中的囚犯得知了,为了不泄露消息,他不得不将所有人带出监狱. 这是一个月黑风高的夜晚...看守们都已经睡着,在没有罪犯打扰的情况下绝对不会醒来,即罪犯不能到达看守所在位置.在每一个空地中,都有一名罪犯,并且同一个空地,能容纳…

地址:http://acm.split.hdu.edu.cn/showproblem.php?pid=3511 题目: Prison Break Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2149    Accepted Submission(s): 681 Problem Description To save Sara, Mi…

1254 - Prison Break   PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB Michael Scofield has just broken out of the prison. Now he wants to go to a certain city for his next unfinished job. As you are the only programmer on h…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3681 前些天花时间看到的题目,但写出不来,弱弱的放弃了.没想到现在学弟居然写出这种代码来,大吃一惊附加敬仰之情.这里借用下他的成果,好好学习吧,骚年*** Sample Input 5 5 GDDSS SSSFS SYGYS SGSYS SSYSS 0 0   Sample Output 4 题意:给出矩阵(作为监狱)和在监狱中的一个装有电池的机器人,其中 F为出发点,图中只有一个,且初始状态下机器…

http://acm.hdu.edu.cn/showproblem.php?pid=3681 题意:一个n*m的矩阵,'F'是起点.机器人从F出发,走到G可以充电,走到Y关掉开关,D不能走进,要求把所有开关关掉,且电量最少,并求出该最小电量. 把F,G,Y的坐标存起来,然后用bfs求出它们每两个点最短距离,然后用dp判断是不是可以满足目的状态. 用二分枚举答案. #include <cstdio> #include <queue> #include <cstring>…

题目大意:n个点m条边的有向图,q次询问c,s,t,表示汽车邮箱容量为c,求从起点s到终点t的最小费用.汽车在每个点可以加任意的油,每个点的单位油价为a[i]. 题目思路:利用最小费优先队列优化最短路,dist[u][use]代表到达u节点,剩余use单位汽油的花费值. #include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> #include<iostream&g…