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//时间复杂度(n+e) n:顶点数 e:边数 #include #include #include #include using 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 #include using 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 #include struct 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 #include struct 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 #include using namespace std; #define MAXN 110000 int 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 Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 37261   Accepted: 10756 Case Time Limit: 2000MS Description You 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): 9209 Problem Description C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 #include using namespace std; #define maxn 1100 struct Edge{ int u,v; }; int pre[maxn],iscut[maxn],bccno[maxn],dfs_clock,bcc_cnt; stack S; vector G[maxn