thoughtspark的博客

#include using namespace std; int pre[1010],sum,n,m,i,j,a,b;; int find(int x){ int r=x; while(pre[r]!=r) r=pre[r]; int z=r,j; //路径压缩 while(z!=r) j=pre[z],pre[z]=r,z=j; //将父亲节点的所有子树全给r return r

1.优先队列 堆的元素个数为N,高度不超过logN+1 节点N的父节点在N/2初，左节点在N*2处 右节点在N*2+1处。 常用代码： #include #include #include using namespace std;   struct cmp1{ booloperator ()(int &a,int &b) { returna>b; }

2.并查集算法以及路径压缩 [置顶] 并查集详解 (转)   来自http://blog.youkuaiyun.com/dellaserss/article/details/7724401>     【转】并查集算法和路径压缩   来自http://www.cnblogs.com/TonyNeal/p/bingchaji.html>   int pre[1000 ]; i

畅通工程续 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 57156    Accepted Submission(s): 21464 Problem Description 某省自从实行了很多年的畅通工程计划后，终于修

选课时间(题目已修改,注意读题) Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5160    Accepted Submission(s): 3970 Problem Description 又到了选课的时