- BC20260025's blog
图论模板合集
- 2023-11-30 17:15:46 @
拓扑排序 B3644
# include<iostream>
# include<cstdio>
# include<vector>
# include<queue>
using namespace std;
const int M=105;
int n,t,d[M];
vector<int> e[M];
queue<int> q;
void toposort(){
for(int i=1;i<=n;++i)
if(d[i]==0) q.push(i);
while(!q.empty()){
int u=q.front();
q.pop();
cout<<u<<" ";
for(int i=0,len=e[u].size();i<len;++i){
int v=e[u][i];
d[v]--;
if(d[v]==0) q.push(v);
}
}
}
int main(){
cin>>n;
for(int i=1;i<=n;++i)
while(true){
cin>>t;
if(t==0) break;
e[i].push_back(t);
d[t]++;
}
toposort();
return 0;
}
单源最短路径模板 P3371 P4779
详见 最短路 合集
最小生成树 P3366
Kruskal
时间复杂度:
# include<iostream>
# include<algorithm>
using namespace std;
const int N=5e3+5,M=2e5+5;
struct edge{
int u,v,w;
void in(){
cin>>u>>v>>w;
}
}e[M];
int n,m,fa[N],k=0,s=0;
bool cmp(edge a,edge b){
if(a.w!=b.w) return a.w<b.w;
else return a.u!=b.u?a.u<b.u:a.v<b.v;
}
int find(int u){
if(u==fa[u]) return u;
return fa[u]=find(fa[u]);
}
void kruskal(){
sort(e,e+m,cmp);
for(int i=1;i<=n;++i) fa[i]=i;
for(int i=0;i<m;++i){
int u=e[i].u,v=e[i].v,w=e[i].w;
int p=find(u),q=find(v);
if(p!=q){
s+=w;++k;
fa[u]=fa[p]=q;
}
if(k==n-1) return;
}
}
int main(){
cin>>n>>m;
for(int i=0;i<m;++i) e[i].in();
kruskal();
if(k==n-1) cout<<s;
else cout<<"orz";
return 0;
}
Prim堆优化
时间复杂度:
# include<iostream>
# include<queue>
# include<vector>
# include<map>
using namespace std;
# define m(a,b) make_pair(a,b)
# define p pair<int,int>
struct edge{
int v,w;
};
const int N=1e5+5,inf=0x7fffffff;
int n,m,x,y,z,sum,k=0;
int dis[N],cnt[N];
bool vis[N];
vector<edge> e[N];
priority_queue<p,vector<p>,greater<p> > q;
void add(int x,int y,int z){
e[x].push_back({y,z});
++cnt[x];
}
void prim(){
for(int i=1;i<=n;++i) dis[i]=inf;
dis[1]=0;
q.push(m(0,1));
while(!q.empty()){
int u=q.top().second,d=q.top().first;
q.pop();
if(vis[u]) continue;
vis[u]=1;
++k;
sum+=d;
if(k>=n) break;
for(int i=0;i<cnt[u];++i){
int v=e[u][i].v,w=e[u][i].w;
if(w<dis[v]){
dis[v]=w;
q.push(m(w,v));
}
}
}
}
int main(){
cin>>n>>m;
for(int i=0;i<m;++i){
cin>>x>>y>>z;
add(x,y,z);
add(y,x,z);
}
prim();
if(k==n) cout<<sum;
else cout<<"orz";
return 0;
}
强连通分量及其应用
详见 强连通分量 合集