Stockbroker GrapevineTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 20558 Accepted: 11141DescriptionStockbrokers are known to overreact to rumours. Y

Sorting It All OutTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 21042 Accepted: 7226DescriptionAn ascending sorted sequence of distinct values is one in which some form of a less-than oper

最近刚学的并查集所以用kruskal来试试最小生成树~kruskal其实用几句话就能说完~ 1.贪心所有边的权值,从小到大取值2.取值时~将边权非0的两个顶点~进行并查操作~如果两个点的祖先不同...边权加入最小生成树...并且将两个点纳入同一个集合中3.判断是否所有点都在同一个集合中完毕~下面上代码~这个代码应该可以作为模版了...但是并查集没有优化~所以复杂度约为0(n

/*很简单的模版题...Dijkstra应该就不用多说了...算法比较简单...尤其是数据结构中的...所以流程图都有...大家一搜一大堆我就不赘述了...赘述也是多余给几组数据3 31 2 11 3 32 3 13 11 2 16 91 2 71 3 91 4 142 3 102 5 153 4 23 5 114 6 95 6 66 81 3 101 5

const int INF = 0x3f3f3f3f; const int MAXN=510;int uN,vN;//u,v数目int g[MAXN][MAXN];//构图int link[MAXN]; //link[v]=u表示右边对左边的匹配bool used[MAXN];//是否访问过bool dfs(int u)//从左边开始找增广路径{ int v; f

/******************Kruskal********************/#include #include #include #include #include using namespace std;const int V = 101;int father[V],map[V][V];struct point { int s,v,rank;}p[V*V

#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std;

#include#include#include#includeusing namespace std;const int N=205;const int INF=0x3f3f3f3f;int s[N][N];//记录图的邻接矩阵int d[N];//记录图中各点的层次int n,m;int min(int a,int b){    return a<b?a:b;}bo

#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std;

const int INF = 0x3fffffff;const int V = 200;int dijkstra(int map[][V],int start,int end,int n) //map为临界矩阵,n为矩阵大小{ int final[V],track[V],i,j,ant=n*n; memset(track,0,sizeof(track)); for(i=1;i<=n

=============================以下是最小生成树+并查集======================================【HDU】1213 How Many Tables 基础并查集★1272 小希的迷宫 基础并查集★1325&&poj1308 Is It A Tree? 基础并查集★1856 More is better

Language:DefaultMPI MaelstromTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 3274 Accepted: 1924DescriptionBIT has recently taken delivery of th

How Many TablesTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6073    Accepted Submission(s): 2894Problem DescriptionToday is I

一开始用邻接矩阵交TLE 分析以后发现 O(n^3) = 1500^3 必须超时但是对于 邻接表 O(m*n) = 1500 * 15000 对于两秒来说完全够了 所以我把木板推倒重写了一遍二分图...还是挺有成就感的...至少以后的题目都可以用把原来低效的木板推掉了都忘记说题意了...就是给你一棵树, 求最小的点能覆盖所有的边...他们说用什么树形DP做我想想貌似也能搞...