当前数据库普遍使用wait-for graph等待图来进行死锁检测 较超时机制,这是一种更主动的死锁检测方式,innodb引擎也采用wait-for graph SQL Server也使用wait-for graph wait-for graph要求数据库保存两种信息 锁的信息链表 事务等待链表 通过上面链表构造出一张图,图中若存在回路,就代表存在死锁,资源间发生相互等待. mysql技术内幕 innodb存储引擎 f f f f
Rikka with Graph Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 190 Accepted Submission(s): 78 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation,
Starting with an undirected graph (the "original graph") with nodes from 0 to N-1, subdivisions are made to some of the edges. The graph is given as follows: edges[k] is a list of integer pairs (i, j, n) such that (i, j) is an edge of the origin
题意:图上的点染色,给出的边的两个点不能都染成黑色,问最多可以染多少黑色. 很水的一题,用dfs回溯即可.先判断和当前点相连的点是否染成黑色,看这一点是否能染黑色,能染色就分染成黑色和白色两种情况递归,如果不能就单递归白色. 代码: #include <cstdio> #include <cstring> const int maxn = 110; int cas, v, e, M; bool g[maxn][maxn]; int color[maxn], rec[maxn]; v
import numpy as np import pandas as pd from matplotlib import pyplot as plt data = pd.DataFrame([[1,2,3],[11,22,33],[111,222,333]]) data.plot() #plot()默认的线性图 #点状随机图 from numpy import random from matplotlib import pyplot def drawScatter(): heights = [
Sparse Graph Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Problem Description In graph theory, the complement of a graph G is a graph H on the same vertices such that two distinct vertices of H are adjacent if