CodeForces - 821D Okabe and City (spfa+思维构图) 题意:给你一个,n*m的二维平面,其中有k个格子被路灯点亮,现在Okabe要从点(1,1)走到点(n,m),他只能走被点亮的格子,而Okabe自己可以点亮一行或者一列的所有格子(在经过后会熄灭且只能在最初亮灯的地方选择点亮一行或者一列的所有格子),问他最少多少次点亮格子可以走到点(n,
题目链接:http://codeforces.com/problemset/problem/144/D思路:首先spfa求出中心点S到其余每个顶点的距离,统计各顶点到中心点的距离为L的点,然后就是要统计在边上的点了,可以枚举边(这里边的数量最多也就100000条),对于枚举的某条边,如果它的其中某个端点到S的距离记过这条边,也就是满足一下这个条件:d1 ...
【codeforces】1063B Labyrinth【SPFA】 1063B Labyrinth 【题目描述】 传送门 【题解】 坑爹题目,BFS是有反例的,如果一条路先到(x,y),但是却不能走到(x1,y1),却阻断了其他路通向这里。所以这题是SPFA,因为你会发现向左走和向右走是相互独立的。也就是说一个只能向左走L步的点和一个只能向右走R步的...
先跑spfa求出最短路构成的DAG,然后在DAG上跑出支配树dfs出size取max即可 关于支配树,因为是DAG,支配点就是入点在支配树上的lca,所以一边拓扑一边预处理倍增,然后用倍增求lca #include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<vector>usingnamespacestd;constintN=400005;intn,m,s,h[N...
最短路专题2 | CodeForces 449B - SPFA算法 编程算法 Jzzhu is the president of country A. There are n cities numbered from 1 to n in his country. City 1 is the capital of A. Also there are m roads connecting the cities. One can go from city to (and vise versa) using the -th road...
Codeforces Round #103 (Div. 2) D. Missile Silos(spfa + 枚举边),题目链接:http://codeforces.com/problemset/problem/144/D思路:首先spfa求出中心点S到其余每个顶点的距离,统计各顶点到中心点的距离为L的点,然后就是要统计在边上的点了,可以枚举边(这里边的数量最
#include <cmath> #include <cctype> using namespace std; #define maxn 200005 #define LL long long int cas=1,T; int n,m,w; const int INF = 0x3f3f3f3f ; struct Edge { int from,to,dist; Edge(){} Edge(int u,int v,int d):from(u),to(v),dist(d){} ...
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不过较大规模的图上SPFA效率确实不高.顺带一个毛子的hack图:http://codeforces.com/blog/entry/3730...
Hello Codeforces! It has been quite some time since I last wandered through this blogs :) I was reading again just for fun about max flow and I remembered a thought I had back when in my training days. It is very common to use a modified version of Bellman-Ford to solve the min cos...