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//时间复杂度(n+e) n:顶点数 e:边数#include#include#include#includeusing namespace std;const int MAXN=10000;struct Node{ int x; int v;};vector Map[MAXN];stack num,T; //T为拓扑序列顶点栈，num为零入度顶点栈i

//时间复杂度(n+e) n:顶点数 e:边数#include#include#includeusing namespace std;const int MAXN=1000;struct cmp{ bool operator()(int x,int y) { return x>y; }};vector map[MAXN];priority_queue,cmp>

//用Treap实现的名次树//名次树支持两个新操作//1: Kth(k):找出第k小元素//2: Rank(x):值x的“名次”，即比x小的结点个数加1#include#includestruct Node{ Node *ch[2]; //左右子树 int r; //优先级。数值越大，优先级越高 int v; //值 i

#include#includestruct Node{ Node *ch[2]; //左右子树 int r; //优先级。数值越大，优先级越高 int v; //值 bool operator <(const Node& rhs)const { //根据优先级比较结点 return r<rhs.r; } int cmp(int x)const{

//倍增算法构造后缀数组//时间复杂度O(nlogn) 空间复杂度O(n)#include#include#includeusing namespace std;#define MAXN 110000int n; //n为处理字符串s的长度char s[MAXN],tmp[MAXN]; //s为输入的字符串int

A Simple Problem with IntegersTime Limit: 5000MS Memory Limit: 131072KTotal Submissions: 37261 Accepted: 10756Case Time Limit: 2000MSDescriptionYou have N i

敌兵布阵Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20994    Accepted Submission(s): 9209Problem DescriptionC国的死对头A国这段时间正在进行军事演习，所以C

//割顶的bccno无意义#include#include#include#includeusing namespace std;#define maxn 1100struct Edge{ int u,v;};int pre[maxn],iscut[maxn],bccno[maxn],dfs_clock,bcc_cnt;stack S;vector G[maxn