262144K   There are NN cities in the country, and MM directional roads from uu to v(1\le u, v\le n)v(1≤u,v≤n). Every road has a distance c_ici​. Haze is a Magical Girl that lives in City 11, she can choose no more than KK roads and make their distanc…

There are N cities in the country, and M directional roads from u to v(1≤u,v≤n). Every road has a distance ci. Haze is a Magical Girl that lives in City 1, she can choose no more than K roads and make their distances become 0. Now she wants to go to…

类似题解 There are NN cities in the country, and MM directional roads from uu to v(1\le u, v\le n)v(1≤u,v≤n). Every road has a distance c_ici​. Haze is a Magical Girl that lives in City 11, she can choose no more than KK roads and make their distances be…

题意:N个点,M条带权有向边,求可以免费K条边权值的情况下,从点1到点N的最短路. 分析:K<=10,用dist[i][j]表示从源点出发到点i,免费j条边的最小花费.在迪杰斯特拉的dfs过程中,每个结点表示的状态有三个属性:访问至的结点,免费的边数和最小花费.将免费的边数看作层,则该图被分为k层.每次更新除了要对边进行松弛操作,也要对层之间进行松弛操作. #include<bits/stdc++.h> using namespace std; typedef long long LL;…

There are NN cities in the country, and MMdirectional roads from uu to v(1\le u, v\le n)v(1≤u,v≤n). Every road has a distance c_ici​. Haze is a Magical Girl that lives in City 11, she can choose no more than KK roads and make their distances become 0…

题目链接:https://nanti.jisuanke.com/t/31001 题目大意:给出一个含有n个点m条边的带权有向图,求1号顶点到n号顶点的最短路,可以使<=k条任意边的权值变为0. 样例输入 复制 1 5 6 1 1 2 2 1 3 4 2 4 3 3 4 1 3 5 6 4 5 2 样例输出 复制 3 解题思路:可以用两种做法,不过都差不多,应该算是同一种思路的不同写法.第一种是在建图时,将一个点拆成k个层次的点,应该总共有k+1层,每个相同层次的点按输入的边权连接,每个点可以向它…

<题目链接> 题目大意: 有N个城市,这些城市之间有M条有向边,每条边有权值,能够选择K条边 边权置为0,求1到N的最短距离. 解题分析: 分层图最短路模板题,将该图看成 K+1 层图,然后具体解析见代码: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <queue> using namespace std;…

题目链接: https://nanti.jisuanke.com/t/31001 超时代码: #include<bits/stdc++.h> using namespace std; # define maxn 10000000+10 # define inf 0x3f3f3f3f int n,m,k; int num; int head[maxn]; int dis[maxn]; int vis[maxn]; struct node { int ne; int to; int co; } q…

https://nanti.jisuanke.com/t/31001 题意 可以把k条边的权值变为0,求s到t的最短路 解析  分层最短路  我们建立k+1层图 层与层之间边权为0,i 向 i+1层转移,代表用了一条免费边. #include <bits/stdc++.h> #define pb push_back #define mp make_pair #define fi first #define se second #define all(a) (a).begin(), (a).en…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 定义dis[i][j]表示到达i这个点. 用掉了j次去除边的机会的最短路. dis[1][0]= 0; 在写松弛条件的时候. 如果用了去除边的机会. 就把机会+1再更新最短路就好. 用spfa会超时. 写个dijkstra+优先队列优化就ok 在dis[n][0..k]中找最小值就好. [代码] #include <bits/stdc++.h> #define LL long long #define ms(a,b) mems…