Warm up 2 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=4619 Description Some 1×2 dominoes are placed on a plane. Each dominoe is placed either horizontally or vertically. It's guaranteed the dominoes in the same direction are not overlapped, but h…

http://acm.hdu.edu.cn/showproblem.php?pid=4619 根据题意可知,每一个方格可能只被一个骨牌覆盖 可能被两个骨牌覆盖 也可能不被覆盖 有一个骨牌覆盖的方格(单覆盖方格)为我们刚开始要找的方格 遍历 每当找到一个单覆盖方格  我们就将这个方格上的骨牌保留 保留这个骨牌的话 有可能会排斥另一个骨牌 如果有排斥的骨牌的话 就将排斥的骨牌去掉 去掉排斥的骨牌 可能产生新的单覆盖方格,则递归下去 最后把所有单覆盖方格处理完  剩下的可能全是双覆盖方格,如果有的话,…

Warm up 2 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 28    Accepted Submission(s): 8 Problem Description Some 1×2 dominoes are placed on a plane. Each dominoe is placed either horizontally…

借用题解上的话,就是乱搞题.. 题意理解错了,其实是坐标系画错了,人家个坐标系,我给当矩阵画,真好反了.对于题目描述和数据不符的问题,果断相信数据了(这是有前车之鉴的hdu 4612 Warm up,明明有重边,出题人欺骗我们的智商)然后就杯具了. 我是用并查集搞得. #include<stdio.h> #include<string.h> ; int mp[MAXN][MAXN]; ]; ]; int find(int x) { return p[x]==x?x:p[x]=fin…

题目:Warm up 2 题意:有横竖两种方式放着的多米诺骨牌,相同方向的不可能重叠,但是横放和竖放             的牌可能重叠.移走重叠的牌使剩下的牌最多. 分析:二分图匹配:最大独立集=顶点数-最大匹配数             横放的为一个点集,竖放的为一个点集. 代码: #include<cstdio> #include<iostream> #include<cstring> using namespace std; struct node { int…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4619 题意: 平面上有一些1×2的骨牌,每张骨牌要么水平放置,要么竖直放置,并且保证同方向放置的骨牌不会相互覆盖.水平放置的牌和竖直放置的牌可能相互覆盖,现在要移去一些牌,使得剩下的牌任何两张都不会相互覆盖,问桌面上最多能剩多少张牌. 分析: 如果把每张牌看作一个结点,则共有两类结点,容易联想到二分图.另外,同方向的牌不会相互覆盖,不同方向的可能相互覆盖,易想到二分图的一个重要性质:同类结点间不会连…

题意:告诉你一些骨牌,然后骨牌的位置与横竖,这样求最多保留多少无覆盖的方格. 这样的话有人用二分匹配,因为两个必定去掉一个,我用的是最小割,因为保证横着和竖着不连通即可. #include <stdio.h> #include <string.h> #include <vector> #include <iostream> #include <queue> #define loop(s,i,n) for(i = s;i < n;i++) u…

三个人整个下午都想不出这题 后来看题解,竟然用匈牙利算法的最大独立集,我顿时晕了. 题意:给竖着和横着的方块,除去重叠的,最多能留下几个方块 #include <cstdlib> #include <iostream> using namespace std; #define N 1010 struct Point{ int x,y; Point(){} Point(int a,int b){x=a;y=b;} bool operator==(const Point &a)…

题目链接 给两种长方形, 水平的和垂直的, 大小都为1*2, n个水平的, m个垂直的, 给出它们的坐标. 水平的和垂直的可以相互覆盖, 但是同种类型的没有覆盖. 去掉一些长方形, 使得剩下的全部都没有覆盖, 求剩下的数量的最大值. 如果一个水平的和一个垂直的相互覆盖, 那么它们之间连一条边, 然后匈牙利匹配求出最大匹配数num, ans = n+m-num; #include <iostream> #include <vector> #include <cstdio>…

给同一张横着的牌的所在的格子编同一样的号,这些格子对应x集合,给同一张竖着的牌所在的格子编同一样的号,对应y集合,同一个格子上既有横着的牌又有竖着的牌,那么就建一条边,有冲突就要拿走一张,结果是总的牌数-最大二分图匹配数··· 贴代码: #include <cstdio> #include <cstring> #define N 1100 int nx,ny; int cx[N],cy[N]; int g[N][N]; bool vis[N]; int map[N][N]; boo…

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; ][]; ][]; ][]; int tmp,ans; //int d[4][2]={{0,-1},{0,1},{1,0},{-1,0}}; int dfs(int i,int j) { ||j<||i>||j>); ][j]) { point1[i+][j]=false; tmp++; dfs(i…

Warm up 2 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 243    Accepted Submission(s): 108 Problem Description Some 1×2 dominoes are placed on a plane. Each dominoe is placed either horizonta…

今天是因为David Lee正好讲这个题的类似题,我才做了一下. 本题是world final 2016的一道水…… 题目地址如下 http://www.lydsy.com/JudgeOnline/problem.php?id=4619. 这道题目让我们求为格式化所有硬盘所需要买的最小的额外空间. 这道题目很容易能看出来是一道贪心题目(据David Lee说还有一种做法是二分+贪心,不过qdez的师天硕stonepage和我Icontofig都觉得二分+贪心有些麻烦,然而stonepage不打算…

Facebook在OSDI 2014上发表论文f4: Facebook's Warm BLOB Storage System,这个系统主要目的就是降低存储成本,在容忍磁盘,主机,机架,数据中心的同时提供2.1倍的存储因子(用户存储的1bit数据实际上占用磁盘2.1bit空间).本文只讨论f4系统的核心Erasure Code部分,如何降低存储因子. Facebook热的blob数据依然存在Haystack中,访问不那么频繁的数据(Warm)放入存储系统f4中.Haystack存储blob的思路就…

Warm up Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 3160    Accepted Submission(s): 718 Problem Description N planets are connected by M bidirectional channels that allow instant transport…

Warm up Problem Description   N planets are connected by M bidirectional channels that allow instant transportation. It's always possible to travel between any two planets through these channels. If we can isolate some planets from others by breaking…

Warm up Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 90    Accepted Submission(s): 12 Problem Description N planets are connected by M bidirectional channels that allow instant transportatio…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4619 一看就知道是二分匹配题目,对每个点拆点建立二分图,最后答案除2.因为这里是稀疏图,用邻接表处理... //STATUS:C++_AC_31MS_480KB #include <functional> #include <algorithm> #include <iostream> //#include <ext/rope> #include <fst…

Warm up Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 5093    Accepted Submission(s): 1131 Problem Description N planets are connected by M bidirectional channels that allow instant transport…

hdu4619:http://acm.hdu.edu.cn/showproblem.php?pid=4619 题意:题目大意:给你两种纸牌 ,一种水平放置共有n张 ,一种竖直放置共有m张.水平放置的纸牌占据点(x, y)和(x + 1 , y) , 竖直放置的纸牌占据点(x , y) 和 (x , y + 1).水平放置的牌之间不会重叠,竖直放置的牌之间也不会重叠,但是水平放置的牌和竖直放置的牌之间可能会重叠.让你拿走一些牌,使剩下的牌之间不会重叠并且数量最多,输出剩余的最大牌数. 题解:对于这…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4619 #include <cstdio> #include <cmath> #include <algorithm> #include <iostream> #include <cstring> #include <queue> #include <vector> #define maxn 1001 #define N 105…

Warm up 2 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 656    Accepted Submission(s): 329 Problem Description Some 1×2 dominoes are placed on a plane. Each dominoe is placed either horizonta…

Warm up Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 1398    Accepted Submission(s): 320 Problem Description N planets are connected by M bidirectional channels that allow instant transport…

题目链接 Warm up Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 5353    Accepted Submission(s): 1195 Problem Description N planets are connected by M bidirectional channels that allow instant tran…

2015年09月06日 17:06:01 yijingjing17 阅读数:9029 标签: PCIEReSet复位Clod resetwarm reset 更多 个人分类: PCIe        PCIe总线规定了两个复位方式:conventional Reset和FLR(FunctionLevel Reset),而Conventional Reset由进一步分为两大类:Fundamental Reset和Non-Fundamental Reset.Fundamental Reset方式包括…

Warm up 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=4612 Description N planets are connected by M bidirectional channels that allow instant transportation. It's always possible to travel between any two planets through these channels. If we can i…

Warm up Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 9073    Accepted Submission(s): 2120 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4612 Description: N planets are connected by M bidire…

Warm up Time Limit:5000MS     Memory Limit:65535KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4612 Description N planets are connected by M bidirectional channels that allow instant transportation. It's always possible to travel bet…

Warm up 虽然网上题解这么多,感觉写下来并不是跟别人竞争访问量的,而是证明自己从前努力过,以后回头复习参考! 题意:n个点由m条无向边连接,求加一条边后桥的最少数量. 思路:如标题,tarjan算法求出所有的桥,然后连通的缩成点,用桥连接这些点,很容易发现这是一颗生成树,我们再加一条边必然成环,要使得桥的数量最少,就得使得这个环中的边最多.于是找这棵树最长的一条链.即树的直径.然后桥的数量减去直径既是答案.为什么不用加一呢,因为加的那条边使得成环不算桥. 总结:此题做了两天,如果全力去做的…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=4612 Warm up Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 7206    Accepted Submission(s): 1681 Problem Description N planets are connecte…