AC，∑ndless

Slim Spanhttp://poj.org/problem?id=3522Time Limit: 5000MSMemory Limit: 65536KDescriptionGiven an undirected weighted graph G, you should find one of spanning trees specified as fol

http://lightoj.com/volume_showproblem.php?problem=1009种类并查集实现（当然用二分图染色也可以）/*0.204s,2820KB*/#includeusing namespace std;const int mx = 20000;int fa[mx * 2 + 5], rk[mx * 2 + 5], x[100005

http://poj.org/problem?id=1182/*235ms,776KB*/#includeconst int mx = 50000 + 5;const int mxadd = 3 * 50000;int fa[mx], rk[mx];int find(int x){ if (x != fa[x]) { int tmp = fa[x]; fa

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3138如何删除？增加一个新结点，标号为所有结点(包括新结点)的最大编号+1然后将旧结点的影响变为0，同时初始化新结点：void delet(int x){ int tx = find(

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=497&page=show_problem&problem=4075由于维护的是树上某一点到根节点的距离，所以可以在find函数中添加一pushdown函数以便更新某点到根的距离。完整代码：/*0.062s*/#

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=497&page=show_problem&problem=3601要安全，这个“图”至多是一颗树（n个点，n-1条边），所以每次判断新加入的两个元素a,b是否都在树上，若是，则加入后树变图（n个点，n条边），此时应拒绝，否则合并a

#includeusing namespace std;const int mx = 100005;int fa[mx], rk[mx];int find(int x) {return fa[x] = (fa[x] == x ? x : find(fa[x]));}void union(int x, int y){ x = find(x), y = find(y); if (

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1549思路：先合并，再遍历每个元素，对其父亲元素计数+1for (i = 1; i <= n; ++i) maxcnt = max(maxcnt, ++cnt[f

http://poj.org/problem?id=1703思路和食物链那题一样的。完整代码：/*297ms,1156KB*/#include#includeint fa[100005], rk[100005];int find(int x){ if (fa[x]) { int tmp = fa[x]; fa[x] = find(fa[x])

http://poj.org/problem?id=2236思路：用set保存已修好的电脑（使用set是为了去重），每次修电脑p时就把set中的与p距离完整代码：/*766ms,640KB*/#include#includeusing namespace std;int x[1005], y[1005];int fa[1005], rk[1005];

Ubiquitous ReligionsTime Limit: 5000MSMemory Limit: 65536KDescriptionThere are so many different religions in the world today that it is difficult to keep track of them all. You are interest

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=33841. 对于第一种命令I p v，我们可以虚拟出一个点Xn = 0，那么p^Xn = v，故第一和第二种命令我们可以统一成p^q = v的模式。2. 记val[i]