dijkstra

啊我沙雕了，竟然以为DJ的邻接矩阵不用初始化。。 #include<bits/stdc++.h> #define R register int using namespace std; //const int inf=0x3f3f3f3f; inline int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9') if(ch=='-') f=-1,ch=getchar(); while(ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar(); return x*f; } int n,m,a[501][501],dis[501],inf; bool vis[501]; inline void Run() { memset(dis,0x3f,sizeof(dis)); inf=dis[1]; dis[1]=0; for(R i=1;i<n;i++) { int x=0; for(R j=1;j<=n;j++) { if((!vis[j])&&(x==0||dis[j]<dis[x])) x=j; } vis[x]=1; for(R k=1;k<=n;k++) //if(a[x][k]) dis[k]=min(dis[k],dis[x]+a[x][k]); } }

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: I am aware of the Dijkstra's shortest path algorithm. However, if I were to modify it so that instead of finding the shortest path it would find the longest path using a greedy algorithm. What would I have to do to the code below: Here is what Im using: as a compare function to select the correct node in the shortest path version: if ( Cost ( potential_node ) > Cost ( current_node ) + cost ( source , current_node )) then cost ( potential_node ) = cost ( current_node ) + cost ( source , current_node ) However, to get to the flip

I have a directed, positive weighted graph. Each edge have a cost of use. I have only A money, i want to calculate shortest paths with dijkstra algorithm, but sum of edges costs on route must be less or equal to A. I want to do this with most smallest Dijstra modification (if I can do it with small modification of Dijkstra). I must do this in O(n*log(n)) if I can, but i think i can. Anyone can help me with this? IVlad https://www.spoj.pl/problems/ROADS/ The problem was given at CEOI '98 and its official solution can be found here . You don't really need to modify Dijkstra's algorithm to do

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have a directed, positive weighted graph. Each edge have a cost of use. I have only A money, i want to calculate shortest paths with dijkstra algorithm, but sum of edges costs on route must be less or equal to A. I want to do this with most smallest Dijstra modification (if I can do it with small modification of Dijkstra). I must do this in O(n*log(n)) if I can, but i think i can. Anyone can help me with this? 回答1: https://www.spoj.pl/problems/ROADS/ The problem was given at CEOI '98 and its official solution can be found here . 回答2: You