HDU4647:Another Graph Game(贪心)】的更多相关文章

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4647 注意这题两人的决策是想要使得自己的分数与对方的差值最大.. 注意到数据范围,显然是贪心之类的,如果没有变那么很简单,如果有边,那么我们进行拆边,把边的权值的一半加到所连的点上.然后排个序贪心.. //STATUS:C++_AC_218MS_1020KB #include <functional> #include <algorithm> #include <iostream…
Problem Description Alice and Bob are playing a game on an undirected graph with n (n is even) nodes and m edges. Every node i has its own weight Wv, and every edge e has its own weight We. They take turns to do the following operations. During each…
题目链接: 题目 Graph Construction Time limit: 3.000 seconds 问题描述 Graph is a collection of edges E and vertices V. Graph has a wide variety of applications in computer. There are different ways to represent graph in computer. It can be represented by adjace…
"Oh, There is a bipartite graph.""Make it Fantastic." X wants to check whether a bipartite graph is a fantastic graph. He has two fantastic numbers, and he wants to let all the degrees to between the two boundaries. You can pick up sev…
https://nanti.jisuanke.com/t/31447 题意 一个二分图,左边N个点,右边M个点,中间K条边,问你是否可以删掉边使得所有点的度数在[L,R]之间 分析 最大流不太会.. 贪心做法: 考虑两个集合A和B,A为L<=d[i]<=R,B为d[i]>R 枚举每个边 1.如果u和v都在B集合,直接删掉2.如果u和v都在A集合,无所谓3.如果u在B,v在A,并且v可删边即d[v]>L4.如果u在A,v在B,并且u可删边即d[u]>L 最后枚举N+M个点判断是…
题目链接 思路题.看的题解. #include <cstdio> #include <string> #include <cstring> #include <algorithm> using namespace std; #define LL __int64 LL p[]; int main() { int i,n,m,sv,ev,w; while(scanf("%d%d",&n,&m)!=EOF) { memset(p…
"Oh, There is a bipartite graph.""Make it Fantastic."X wants to check whether a bipartite graph is a fantastic graph. He has two fantastic numbers, and he wants to let all the degrees to between the two boundaries. You can pick up seve…
E. Pashmak and Graph time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Pashmak's homework is a problem about graphs. Although he always tries to do his homework completely, he can't solve thi…
生物信息学 Contig是reads拼成的连续的DNA片段,连续表达一个gene.通过双端测序的contig可确定contig之间的关系得到scaffold,Scaffold是reads拼成的有gap的DNA片段.理想情况下,一条染色体用同一个scaffold的表达.整个genome存在很多零碎片段,可舍弃.因为duplication产生很多overlap. N50,L50和NG50是评价genome assembly的quality的标准,评价长度时使用N50,N50是一个contig的长度.…
LINK:Giant Graph 神仙题目. 容易发现在图中选择某个点的贡献为\(10^{18\cdot(x+y+z)}\) 这等价于多选一个点多大一点就多乘了一个\(10^{18}\) 所以显然是贪心的选取是最优的. 直接贪复杂度较高 考虑一个点的是否选取只和其某个维度上相邻的点有关. 形式化的 设\(f_{i,j,k}\)表示当前这个点是否选择 那么有\(f_{i,j,k}=\Pi [f_{i',j'.k'}=0]\) 可以观察出来这是一个DAG.接下来的一步就比较神仙了. 这个dp转移每次…