//POJ2679 //DFS+SPFA+邻接表 //只能走每个点费用最小的边,相同则需保证距离最短 //求最小费用及最短距离 //Time:47Ms Memory:900K #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<queue> using namespace std; #define MAXN 1105 #define…

transaction transaction transaction Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 1496    Accepted Submission(s): 723 Problem Description Kelukin is a businessman. Every day, he travels arou…

判断是欧拉通路后,DFS简单剪枝求解字典序最小的欧拉通路路径 //Time:16Ms Memory:228K #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; #define MAX 1005 #define MAXS 24 //姓名 #define MAXN 26 //字母 struct Edge{ char name…

本题可以通过全部n位二进制数作点,而后可按照某点A的末位数与某点B的首位数相等来建立A->B有向边,以此构图,改有向图则是一个有向欧拉回路,以下我利用DFS暴力求解该欧拉回路得到的字典序最小的路径. //求咬尾数,一个2^n位环形二进制数,该二进制的每n位连续二进制数都不同 //DFS求解欧拉回路 //Time:32ms Memory:1668K #include<iostream> #include<cstring> #include<cstdio> using…

//转移为最短路问题,枚举必经每一个不小于酋长等级的人的最短路 //Time:16Ms Memory:208K #include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<algorithm> using namespace std; #define INF 0x3f3f3f3f #define MAX 105 int lim, n; int p[M…

通过几道例题简单阐述一下DFS的相关题型 ZOJ2412-Farm Irrigation 直观的DFS题型,稍加变化,记录好四个方向上的通路就能够做出来 题目和接水管类似,问最少要灌溉几次,即求解最少有多少个连通子图. //和接水管游戏类似,将相应水管通路标记清晰即可 //Time:0Ms Memory:270K #include<iostream> #include<cstring> #include<cstdio> using namespace std; #def…

Sparse Graph Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 689    Accepted Submission(s): 238 Problem Description In graph theory, the complement of a graph G is a graph H on the same vertic…

transaction transaction transaction Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 1496    Accepted Submission(s): 723 Problem Description Kelukin is a businessman. Every day, he travels arou…

AcWing 851 spfa求最短路 题解 以此题为例介绍一下图论中的最短路算法 \(Bellman\)-\(Ford\) 算法.算法的步骤和正确性证明参考文章最短路径(Bellman-Ford算法) 松弛函数 对边集合 \(E\) 中任意边,\(w(u,v)\) 表示顶点 \(u\) 到顶点 \(v\) 的边的权值,用 \(d[v]\) 表示当前从起点 \(s\) 出发到顶点 \(v\) 的最短距离. 若存在边 \(e\),权值为 \(w(u,v)\),使得: \[d[v] > d[u] +…

Mart Master II Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 675    Accepted Submission(s): 237 Problem Description Trader Dogy lives in city S, which consists of n districts. There are n - 1…