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POJ 3580 —— Splay模板
SuperMemoTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 7161 Accepted: 2368Case Time Limit: 2000MSDescriptionYour friend, Jackson is invited to a原创 2013-10-02 17:36:54 · 802 阅读 · 0 评论 -
USACO 4.2.2 && POJ 1274 —— 二分图匹配
The Perfect StallTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 16733 Accepted: 7659DescriptionFarmer John completed his new barn just last week, compl原创 2013-10-31 03:44:05 · 818 阅读 · 0 评论 -
POJ 2680 —— 最小费用流求解区间图的最大权独立集问题
IntervalsTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 5967 Accepted: 2400DescriptionYou are given N weighted open intervals. The ith interval covers原创 2013-10-31 02:26:21 · 1220 阅读 · 0 评论 -
POJ 1127 —— 计算几何 && 线段相交
Jack StrawsTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 2860 Accepted: 1294DescriptionIn the game of Jack Straws, a number of plastic or wooden "stra原创 2013-10-22 20:01:46 · 970 阅读 · 0 评论 -
POJ 2135 —— 最小费用流
Farm TourTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9786 Accepted: 3603DescriptionWhen FJ's friends visit him on the farm, he likes to show them ar原创 2013-10-30 01:21:26 · 794 阅读 · 0 评论 -
POJ 3469 —— 最大流
Dual Core CPUTime Limit: 15000MS Memory Limit: 131072KTotal Submissions: 17034 Accepted: 7350Case Time Limit: 5000MSDescriptionAs more and more computers ar原创 2013-10-29 16:00:51 · 765 阅读 · 0 评论 -
POJ 3281 —— 最大流
DiningTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 7987 Accepted: 3644DescriptionCows are such finicky eaters. Each cow has a preference for certain原创 2013-10-29 15:07:22 · 691 阅读 · 0 评论 -
POJ 2186 —— 分解强连通分量
Popular CowsTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 20933 Accepted: 8530DescriptionEvery cow's dream is to become the most popular cow in the he原创 2013-10-25 01:36:08 · 736 阅读 · 0 评论 -
POJ 2187 —— 凸包 + 旋转卡壳 求多边形的直径
Beauty ContestTime Limit: 3000MS Memory Limit: 65536KTotal Submissions: 25248 Accepted: 7752DescriptionBessie, Farmer John's prize cow, has just won first place i原创 2013-10-24 20:19:04 · 829 阅读 · 0 评论 -
堆的数组实现
///数组表示二叉树int heap[MAXN] , sz = 0;void push(int x){ int i = sz++;///自己节点的编号 while(i > 0) { ///父亲节点的编号 int p = (i - 1) / 2; ///如果没有大小颠倒则退出 if(heap[p] <= x原创 2013-10-10 19:07:54 · 828 阅读 · 0 评论 -
并查集
时间复杂度O(a(n)),a(n)为阿克曼函数的反函数,比log(n)还快。科普一下阿克曼函数http://zh.wikipedia.org/wiki/%E9%98%BF%E5%85%8B%E6%9B%BC%E5%87%BD%E6%95%B8int par[MAXN];int rank[MAXN];void init(int n){ for(int i = 0 ; i原创 2013-10-10 22:24:48 · 740 阅读 · 0 评论 -
二叉搜索树的三种实现 —— 指针 | set | map
一、指针实现:///表示节点的结构体struct node{ int val; node *lch , *rch;};///插入数值xnode *insert(node *p , int x){ if(p == NULL) { node *q = new node; q -> val = x; q原创 2013-10-10 21:25:00 · 1146 阅读 · 0 评论 -
HDOJ 1068 —— 最大独立集
Girls and BoysTime Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6250 Accepted Submission(s): 2797Problem Descriptionthe secon原创 2013-11-11 05:00:43 · 969 阅读 · 0 评论