Rescue Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 19985    Accepted Submission(s): 7110 Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is d…

挺有意思的一道题目,解法是AC自动机+DP.AC自动机建立fail指针时,一定要注意结点的属性也需要传递.AC自动机结合了trie和kmp的优点.需要注意的是,每个模式串仅计算一次,否则这题很难解. /* 4057 */ #include <iostream> #include <sstream> #include <string> #include <map> #include <queue> #include <set> #inc…

BFS+优先级队列. #include <iostream> #include <cstdio> #include <cstring> #include <queue> using namespace std; #define MAXNUM 205 typedef struct node_st { int x, y ,t; node_st() {} node_st(int xx, int yy, int tt) { x = xx; y = yy; t = t…

Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison. Angel's friends want to save Angel. Their task is: approa…

题目传送门 题意:从r走到a,遇到x多走一步,问最小走到a的步数 分析:因为r有多个,反过来想从a走到某个r的最小步数,简单的BFS.我对这题有特殊的感情,去年刚来集训队时肉鸽推荐了这题,当时什么都不会,看个数组模拟队列的BFS看的头晕,现在看起来也不过如此,额,当年开始是从r走到a的,因为数据巨弱才过的,应该要用到优先队列. /************************************************ * Author :Running_Time * Created Ti…

Rescue Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 31648    Accepted Submission(s): 11083 Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 56784    Accepted Submission(s): 19009 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3049    Accepted Submission(s): 2364 Problem Description You are the CEO of Nasty Hacks Inc., a company that creates small pieces of…

Box of Bricks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5994    Accepted Submission(s): 2599 Problem Description Little Bob likes playing with his box of bricks. He puts the bricks one up…

本节我们讨论 Suspend/Resume 和 Rescue/Unrescue 这两组操作. Suspend/Resume 有时需要长时间暂停 instance,可以通过 Suspend 操作将 instance 的状态保存到宿主机的磁盘上.当需要恢复的时候,执行 Resume 操作,从磁盘读回 instance 的状态,使之继续运行. 这里需要对 Suspend 和 Pause 操作做个比较: 相同点两者都是暂停 instance 的运行,并保存当前状态,之后可以通过 Resume 操作恢复…