摘要： A. Sereja and Coat Racktime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputSereja owns a restaurant fornpeople. The restaurant hall has a coat rack withnhooks. Each restaurant visitor can use a hook to hang his clothes on it. Using thei-th hook costsa 阅读全文

摘要： #include #include #include #include #include int sum;int n,m;int a[100010];int l[100010];int s[100010];int map[100010];int main(){ while(~scanf("%d%d",&n,&m)) { memset(a,0,sizeof(a)); memset(l,0,sizeof(l)); memset(s,0,sizeof(s));; for(int i = 1;i = 1;i --) { ... 阅读全文

摘要： Hard Disk DriveTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 125Accepted Submission(s): 75 Problem Description Yesterday your dear cousin Coach Pang gave you a new 100MB hard disk drive (HDD) as a gift because you will get married next year. ... 阅读全文

摘要： find your present (2)Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/1024 K (Java/Others) Total Submission(s): 13413Accepted Submission(s): 5049 Problem DescriptionIn the new year party, everybody will get a "special present".Now it's your turn to get your special present, a 阅读全文

摘要： 连通性问题主要有求有向图的强连通分量的三种算法，Kosaraju，Trajan，Carbow:最小点基和最小权点基础：无向图的双连通分量以及求桥和割点的Trarjan算法：全局最小权割问题和Stoer—Wagner算法：2—SAT本来图论的学习在十一月份的时候我应该结束第一轮的，但是由于该死的考试，第一轮只能进行到无向图的双连通了，图论还剩下网络流和二分图及匹配算法，两个大的专题，没有办法只能在来年一，二月份进行了，本来那段时间是用来图论进阶的和学习计算几何的，哎，坑爹的考试，可不能挂科啊，啊Ksaraju：Ksaraju算法基于强连通分量内部所有的点都可以互达的，那么对原图G和反图GT分别进 阅读全文

摘要： 点连通度与边连通度在一个无向连通图中，如果有一个顶点集合，删除这个顶点集合，以及这个集合中所有顶点相关联的边以后，原图变成多个连通块，就称这个点集为割点集合。一个图的点连通度的定义为，最小割点集合中的顶点数。类似的，如果有一个边集合，删除这个边集合以后，原图变成多个连通块，就称这个点集为割边集合。一个图的边连通度的定义为，最小割边集合中的边数。[双连通图、割点与桥]如果一个无向连通图的点连通度大于1，则称该图是点双连通的(pointbiconnected)，简称双连通或重连通。一个图有割点，当且仅当这个图的点连通度为1，则割点集合的唯一元素被称为割点(cutpoint)，又叫关节点(artic 阅读全文

摘要： Count the rectanglesTimeLimit:2 Second MemoryLimit:64 MegabyteTotalsubmit:64 Accepted:36 DescriptionYou are given numbers of rectangles made of '1's and '0's.Calculate how many small rectangles which have four '1's on the four corners.InputThere are several test cases, each t 阅读全文

摘要： PhotoTimeLimit:2 Second MemoryLimit:64 MegabyteTotalsubmit:33 Accepted:13 DescriptionFJ wants to take pictures of his N cows (2 #include #include #include #include using namespace std;int can_be[1010][1010];int map[10010];int the_last_sum;int n,k;int a,b;struct Node{ int x; int posi;}v[10010];in... 阅读全文

摘要： Breed ProximityTimeLimit:2 Second MemoryLimit:64 MegabyteTotalsubmit:46 Accepted:10 DescriptionFarmer John's N cows (1 #include #include #include #include using namespace std;int map[51010];int can_be[1100000];int the_last_max;int n,k;int main(){ while(~scanf("%d%d",&n,&k)) { m 阅读全文

摘要： Graph’s Cycle ComponentTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 1655Accepted Submission(s): 637 Problem DescriptionIn graph theory, a cycle graph is an undirected graph that consists of a single cycle, or in other words, some number of ver 阅读全文