题目大意:同这道题,只是把最大值变成了最小值 题解:略 卡点:无 C++ Code: #include <cstdio> #define maxn 100010 #define maxm 100010 int head[maxn], cnt; struct Edge { int to, nxt; long long w; } e[maxm << 1]; void addE(int a, int b, long long c) { e[++cnt] = (Edge) {b, head…

Shortest Path Problem? You are given an undirected graph with weighted edges. The length of some path between two vertices is the bitwise xor of weights of all edges belonging to this path (if some edge is traversed more than once, then it is include…

00 前言 各位小伙伴大家好,相信大家已经看过前面column generation求解vehicle routing problems的过程详解.该问题中,子问题主要是找到一条reduced cost最小的合法路径,然后加入到Master Problem中.其实,子问题也是一个著名的NP-Hard问题,今天我们就来介绍一下. 01 ESPPRC 考虑图1.1中描述的网络. 除了每条边的成本c_ij之外,还存在经过边(i,j)的所消耗的资源t_ij,比如时间. 我们的目标是找到从开始节点到结束节…

Description 给定一张 \(n\) 个点 \(m\) 条边的无向图,一开始你在点 \(1\),且价值为 \(0\) 每次你可以选择一个相邻的点,然后走过去,并将价值异或上该边权 如果在点 \(n\),你可以选择结束游戏 求一种方案,使得结束游戏后价值最小 \(n,m \le 10^5\) Input 第一行为两个整数\(n,m\)代表有\(n\)个点\(m\)条边. 接下来\(m\)行,描述一条从\(x\)到\(y\)长度为\(z\)的无向边. Output 一个整数,代表最小价值.…

time limit per test 3 seconds memory limit per test 512 megabytes input standard input output standard output You are given an undirected graph with weighted edges. The length of some path between two vertices is the bitwise xor of weights of all edg…

http://codeforces.com/problemset/problem/845/G 从顶点1dfs全图,遇到环则增加一种备选方案,环上的环不需要走到前一个环上作为条件,因为走完第二个环可以从第一个环原路返回. 对每种备选方案通过x = min(x,x^v[i]) 保留备选方案中不存在的最高位,如果能由已经存在的备选方案组合得到则结果为零 最后对val[n]做一次类似的操作得到答案 #include<bits/stdc++.h> using namespace std; ; vecto…

因为每行只有一个区域不能往下走,因此我们可以来分析一下从起点到整个矩形每个位置的最短路.可以发现每一行的最短路只与上一行的最短路有关,假设我们知道上一行的最短路,上一行不能往下走的区间在 \([L, R]\),那么可以发现的是 \([1, L - 1], [R + 1, m]\) 这些区间会直接从上一行走下来,因为假设存在一个更靠前的位置它走过来更优,那么前一行的最短路就会由这个位置走过来更新.同理,因为区间 \([L, R]\) 是不能从上面直接走下来的,所以其一定是从 \([1, L - 1…

We know that the longest path problem for general case belongs to the NP-hard category, so there is no known polynomial time solution for it. However, for a special case which is directed acyclic graph, we can solve this problem in linear time. First…

We all know that the shortest path problem has optimal substructure. The reasoning is like below: Supppose we have a path p from node u to v, another node t lies on path p: u->t->v ("->" means a path). We claim that u->t is also a sh…

The Shortest Path Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2440    Accepted Submission(s): 784 Problem Description There are N cities in the country. Each city is represent by a matrix si…