Doing Homework Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12145    Accepted Submission(s): 5851 Problem Description Ignatius has just come back school from the 30th ACM/ICPC. Now he has a l…

Doing Homework Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14600    Accepted Submission(s): 7070 Problem Description Ignatius has just come back school from the 30th ACM/ICPC. Now he has a l…

由于数据量较小,直接二进制模拟全排列过程,进行DP,思路由kuangbin blog得到,膜拜斌神 #include<iostream> #include<cstdio> #include<cstdlib> #include<cmath> #include<algorithm> #include<cstring> using namespace std; struct asd { ]; int deadline,cost; } o[]…

Resource Archiver Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others)Total Submission(s): 2382    Accepted Submission(s): 750 Problem Description Great! Your new software is almost finished! The only thing left to…

Wireless Password Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5640    Accepted Submission(s): 1785 Problem Description Liyuan lives in a old apartment. One day, he suddenly found that there…

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1074 题意:给你n个课程(n<=15)每个课程有限制的期限和完成该课程的时间,如果超出时间,每超一天扣一分,问完成全部课程,最少会扣多少分. 题解:典型的状压DP #include<cstdio> #define FFC(i,a,b) for(int i=a;i<=b;++i) ]; ];]; <<];//mi表示最小扣分,pre表示上一种状态,d表示当前时间 void o…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5765 [题目大意] 给出一张图,求每条边在所有边割集中出现的次数. [题解] 利用状压DP,计算不同的连通块,对于每条边,求出两边的联通块的划分方案数,就是对于该点的答案. [代码] #include <cstdio> #include <algorithm> #include <cstring> using namespace std; int n,m,T,Cas=1…

Problem Description Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will r…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3681 思路:机器人从出发点出发要求走过所有的Y,因为点很少,所以就能想到经典的TSP问题.首先bfs预处理出‘Y',’F','G'之间的最短距离,由于G点可以充电,到达G点就把当前能量更新为电池容量然后继续走.因为每个G点只能充一次电,这就好像TSP中的每个点只能走一次一样,然后就是二分答案了,用状压DP判定当前电池容量的情况下是否能符合条件. #include<iostream> #includ…

转自wdd :http://blog.csdn.net/u010535824/article/details/38540835 题目链接:hdu 4778 状压DP 用DP[i]表示从i状态选到结束得到的最大值 代码也来自wdd /****************************************************** * File Name: b.cpp * Author: kojimai * Creater Time:2014年08月13日 星期三 11时42分53秒 *…