var n,m,s,t,v,i,a,b,c:longint;//这道题的代码不是这个,在下面 first,tr,p,q:..]of longint; next,eb,ew:..]of longint; procedure swap(a,b:longint); var t:longint; begin t:=tr[a];tr[a]:=tr[b];tr[b]:=t; t:=p[a];p[a]:=p[b];p[b]:=t; t:=q[p[a]];q[p[a]]:=q[p[b]];q[p[b]]:=t;
做这道题的动机就是想练习一下堆的应用,顺便补一下好久没看的图论算法. Dijkstra算法概述 //从0出发的单源最短路 dis[][] = {INF} ReadMap(dis); for i = 0 -> n - 1 d[i] = dis[0][i] while u = GetNearest(1 .. n - 1, !been[]) been[u] = 1 for_each edge from u d[edge.v] = min(d[edge.v], d[u] + dis[u][edge.v]
#include<algorithm> #include<iostream> #include<cstdio> #include<cstring> #include<queue> #define M 100000 #define pa pair<int,int>//优先比较第一个元素 using namespace std; int d[M],n,m,cnt,head[M],next[M],u[M],dis[M],num,s,t; b
You are given a list of cities. Each direct connection between two cities has its transportation cost (an integer bigger than 0). The goal is to find the paths of minimum cost between pairs of cities. Assume that the cost of each path (which is the s
Problem UVA - 11374 - Airport Express Time Limit: 1000 mSec Problem Description In a small city called Iokh, a train service, Airport-Express, takes residents to the airport more quickly than other transports. There are two types of trains in Airp
Til the Cows Come Home Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.