Going from u to v or from v to u? Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 17993 Accepted: 4816 Description In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridor
//拓扑排序判断是否有环 #include<cstdio> #include<algorithm> #include<string.h> #include<math.h> #include<queue> using namespace std; typedef long long ll; ; int G[maxn][maxn]; int in[maxn]; void init() { memset(G,,sizeof(G)); //图 memse
拓扑序列的判断方法为不存在有向环,代码实现的话有两种,一种是直接去判断是否存在环,较为难理解一些,另一种的话去判断结点入度,如果存在的入度为0的点大于一个,则该有向图肯定不存在一个确定的拓扑序列 #include<cstdio> #include<cstring> #include<iostream> #include<cstdlib> using namespace std; ]; int m,n,u,v; ][]; int dfs(int u) { vi
分析:BFS判断是否有环. #include<bits/stdc++.h> using namespace std; typedef long long ll; int gra[200][200]; int vis[100]; void bfs(int n) { memset(vis,0,sizeof(vis)); vis[1] = 1; int q[100]; int in = 0,out = 0, f = 0; q[in ++] = 1; while(in > out) { int
public static boolean isBSTSequence(int[] s,int l, int r) { if (s == null || r <= 0) return false; int n = r; int root = s[n - 1]; int i = 0; for (; i < n - 1; i++) { if (s[i] > root) break; } int j = i; for (; j < n - 1; j++) { if (s[j] <
[BZOJ3012][Usaco2012 Dec]First! Description Bessie has been playing with strings again. She found that by changing the order of the alphabet she could make some strings come before all the others lexicographically (dictionary ordering). For instance