bianyanze @ 2024-10-15 19:51:08
帮助者互关 我快考CSP_J复赛了,谢谢帮助! 献上本人代码:
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
const int M=5*1e5+10;
int dis[N],n,m,vis[N],s;
int head[M],tot;
struct node{
int to,nxt,w;
}edge[M];
int u,v,w;
void add(int u,int v,int w){
edge[tot++]=(node){
v,head[u],w
};
head[u]=tot;
}
priority_queue<pair<int,int> >q;
void dijkstra(int s){
q.push(make_pair(0,s));
dis[s]=0;
while(!q.empty()){
int u=q.top().second;
q.pop();
if(vis[u])continue;
vis[u]=1;
for(int i=head[u];i;i=edge[i].nxt){
int v=edge[i].to,w=edge[i].w;
if(dis[v]>dis[u]+w){
dis[v]=dis[u]+w;
if(!vis[v])q.push(make_pair(-dis[v],v));
}
}
}
}
int main(){
scanf("%d%d%d",&n,&m,&s);
for(int i=1;i<=n;i++){
dis[i]=2147483647;
}
for(int i=1;i<=m;++i){
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
}
dijkstra(s);
for(int i=1;i<=n;i++){
cout<<dis[i]<<" ";
}
return 0;
}
by Jacky29 @ 2024-10-15 19:52:02
物理意义的在旁边
by J20220912 @ 2024-10-17 09:02:23
void add(int u,int v,int w){
edge[tot++]=(node){
v,head[u],w
};
head[u]=tot;
}
改成
void add(int u,int v,int w){
edge[++tot]=(node){
v,head[u],w
};
head[u]=tot;
}
by J20220912 @ 2024-10-17 19:08:37
@bianyanze
by bianyanze @ 2024-10-19 19:42:03
why?
by sgxsz @ 2024-10-24 20:32:50
邻接表存边的编号要从1开始 如果是涉及双向边的x^1运算还要从2开始
for(int i=head[u];i;i=edge[i].nxt)
你的i就是再枚举编号,0终止 如果不想改tot其实也行,但要改for