维护环内最大值与严格次大值
与未放入最小生成树的边枚举加入
#include#define re return#define dec(i,l,r) for(ll i=l;i>=r;--i)#define inc(i,l,r) for(ll i=l;i<=r;++i)typedef long long ll;using namespace std;template inline void rd(T&x){ char c;bool f=0; while((c=getchar())<'0'||c>'9')if(c=='-')f=1; x=c^48; while((c=getchar())>='0'&&c<='9')x=x*10+(c^48); if(f)x=-x;}const int maxn=100005,maxm=300005;ll n,m,k=1,hd[maxn],deep[maxn];ll max1[maxn][21],max2[maxn][21],fa[maxn][21];ll vis[maxm],f[maxn];ll sum,ans=999999999999999999;struct node{ ll fr,to,nt,val; bool operator<(node p)const { re val max1[fa[x][i]][i]) { max1[x][i+1]=max1[x][i]; max2[x][i+1]=max(max1[fa[x][i]][i],max2[x][i]); } else { max1[x][i+1]=max1[fa[x][i]][i]; max2[x][i+1]=max(max1[x][i],max2[fa[x][i]][i]); } } for(ll i=hd[x];i;i=e1[i].nt) { ll v=e1[i].to; if(v==fa[x][0])continue; fa[v][0]=x; max1[v][0]=e1[i].val; max2[v][0]=-2147483647; dfs(v); }}inline ll find(ll x){ re x==f[x]?x:f[x]=find(f[x]);}int main(){ // freopen("in.txt","r",stdin);freopen("tree8.in","r",stdin); ll x,y,z; rd(n),rd(m); inc(i,1,n)f[i]=i; inc(i,1,m) { rd(x),rd(y),rd(z); e[i]=(node){x,y,0,z}; } sort(e+1,e+m+1); k=0; ll cnt=0; inc(i,1,m) { x=find(e[i].to),y=find(e[i].fr); if(x!=y) { add1(e[i].to,e[i].fr,e[i].val); sum+=e[i].val; vis[i]=1; f[x]=y; ++cnt; if(cnt==n-1)break; } } dfs(1); inc(j,1,m) if(!vis[j]) { x=e[j].to;y=e[j].fr;//-------------------------------------------------------------------------------- ll d1=0,d2=0; if(deep[x] =deep[y]) { if(max1[x][i]>d1) { d2=max(d1,max2[x][i]); d1=max1[x][i]; } else if(max1[x][i]!=d1) d2=max(d2,max1[x][i]); x=fa[x][i]; } if(x!=y) { dec(i,20,0) if(fa[x][i]!=fa[y][i]) { if(max1[x][i]>d1) { d2=max(d1,max2[x][i]); d1=max1[x][i]; } else if(max1[x][i]!=d1) d2=max(d2,max1[x][i]); if(max1[y][i]>d1) { d2=max(d1,max2[y][i]); d1=max1[y][i]; } else if(max1[y][i]!=d1) d2=max(d2,max1[y][i]); x=fa[x][i];y=fa[y][i]; } if(max1[x][0]>d1) { d2=d1; d1=max1[x][0]; } else if(max1[x][0]!=d1) d2=max(d2,max1[x][0]); if(max1[y][0]>d1) { d2=d1; d1=max1[y][0]; } else if(max1[y][0]!=d1) d2=max(d2,max1[y][0]); } //----------------------------------------------------------------------------------------------------- if(d1==e[j].val)ans=min(ans,sum+e[j].val-d2); else ans=min(ans,sum+e[j].val-d1); } printf("%lld",ans); re 0;}