danlao @ 2024-05-04 16:35:49
#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize(2)
//#pragma GCC optimize(3, "Ofast", "inline")
#define hh putchar('\n')
#define kg putchar(' ')
#define debug puts("debug")
//#define int long long
//#define int __int128
namespace quickread{
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;
}
inline void write(int x){
if(x<0){putchar('-');x=-x;}
if(x>9)write(x/10);
putchar(x%10+'0');
}
inline string readstr(){
char ch=getchar();string str="";
while(!(ch>='0'&&ch<='9')&&!(ch>='a'&&ch<='z')&&!(ch>='A'&&ch<='Z')) ch=getchar();
while((ch>='0'&&ch<='9')||(ch>='a'&&ch<='z')||(ch>='A'&&ch<='Z')){str+=ch;ch=getchar();}
return str;
}
inline char readchar(){
char ch=getchar();
while(!(ch>='0'&&ch<='9')&&!(ch>='a'&&ch<='z')&&!(ch>='A'&&ch<='Z')) ch=getchar();
return ch;
}
}
using namespace quickread;
const int N=1e6+10;
int n,m,dis[N],ans;
bool vis[N];
struct edge{
int v,w;
};
vector<edge>p[N],g[N];
struct csp{
int u,w;
};
bool operator > (const csp& a,const csp& b){
return a.w>b.w;
}
void dijkstra_p(int u){
memset(dis,0x3f,sizeof(dis));memset(vis,0,sizeof(vis));
dis[u]=0;
priority_queue<csp ,vector<csp> ,greater<csp> >q;
q.push(csp{u,0});
while(!q.empty()){
u=q.top().u;
q.pop();
if(vis[u]) continue;
vis[u]=1;
for(edge to:p[u]){
int v=to.v,w=dis[u]+to.w;
if(dis[v]>w){
dis[v]=w;
q.push(csp{v,w});
}
}
}
for(int i=2;i<=n;i++)
ans+=dis[i];
}
void dijkstra_g(int u){
memset(dis,0x3f,sizeof(dis));memset(vis,0,sizeof(vis));
dis[u]=0;
priority_queue<csp ,vector<csp> ,greater<csp> >q;
q.push(csp{u,0});
while(!q.empty()){
u=q.top().u;
q.pop();
if(vis[u]) continue;
vis[u]=1;
for(edge to:g[u]){
int v=to.v,w=dis[u]+to.w;
if(dis[v]>w){
dis[v]=w;
q.push(csp{v,w});
}
}
}
for(int i=2;i<=n;i++)
ans+=dis[i];
}
signed main(){
// ios::sync_with_stdio(0);
// cin.tie(0);cout.tie(0);
n=read(),m=read();
for(int i=1;i<=n;i++){
int u=read(),v=read(),w=read();
p[u].push_back(edge{v,w});
g[v].push_back(edge{u,w});
}
dijkstra_p(1);
dijkstra_g(1);
write(ans);
return 0;
}
by Kindershiuo @ 2024-06-24 14:16:18
方向建边跑一遍dijkstra
正向建边再跑一遍dijkstra就好了
by Kindershiuo @ 2024-06-24 14:16:34
@Kindershiuo 反向