#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#define Max(a,b) (a>b?a:b)
#define N 100005
#define INF 0x3f3f3f3f
using namespace std;
int read()
{int x=0,f=1;char c=getchar();while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}return x*f;
}
int n,m,a[N],l[303],r[303];
int seg[606],ans=0;
vector<int>A,cp;
struct Node
{int l,r,maxn,lazy;
}t[N*4];
void build(int idx,int l,int r)
{t[idx].l=l,t[idx].r=r;if(l==r){t[idx].maxn=a[l],t[idx].lazy=0;return;}int mid=(l+r)>>1;build(idx<<1,l,mid),build(idx<<1|1,mid+1,r);t[idx].maxn=Max(t[idx<<1].maxn,t[idx<<1|1].maxn);
}
void pushdown(int idx)
{if(t[idx].lazy&&t[idx].l!=t[idx].r){int lz=t[idx].lazy;t[idx].lazy=0;t[idx<<1].lazy+=lz,t[idx<<1|1].lazy+=lz;t[idx<<1].maxn+=lz,t[idx<<1|1].maxn+=lz;}
}
void add(int idx,int l,int r,int w)
{if(l<=t[idx].l&&r>=t[idx].r){t[idx].maxn+=w,t[idx].lazy+=w;return;}pushdown(idx);int mid=(t[idx].l+t[idx].r)>>1;if(r<=mid)add(idx<<1,l,r,w);else if(l>mid)add(idx<<1|1,l,r,w);else add(idx<<1,l,r,w),add(idx<<1|1,l,r,w);t[idx].maxn=Max(t[idx<<1].maxn,t[idx<<1|1].maxn);
}
int main()
{n=read(),m=read();for(int i=1;i<=n;i++)a[i]=read();build(1,1,n);for(int i=1;i<=m;i++){l[i]=read(),r[i]=read();seg[i*2-1]=l[i],seg[i*2]=r[i];}sort(seg+1,seg+2*m+1),seg[0]=0;int cnt=unique(seg+1,seg+2*m+1)-seg-1;seg[cnt+1]=INF;int now=0,f;for(int i=1;i<=n;i++){if(i==seg[now]||(i>seg[now]&&i==seg[now+1])){f=0,cp.clear();if(i==seg[now+1])now++;for(int j=1;j<=m;j++){if(l[j]<=i&&r[j]>=i)f++,add(1,l[j],r[j],-1),cp.push_back(j);}}else if(i>seg[now]){now++,f=0,cp.clear();for(int j=1;j<=m;j++){if(l[j]<=seg[now-1]+1&&r[j]>=seg[now]-1)f++,add(1,l[j],r[j],-1),cp.push_back(j);}}int minnum=a[i]-f;if(t[1].maxn-minnum>ans){ans=t[1].maxn-minnum;A=cp;}if(i==seg[now]-1||i==seg[now]){for(int j=0;j<cp.size();j++)add(1,l[cp[j]],r[cp[j]],1);}}printf("%d\n%d\n",ans,A.size());for(int i=0;i<A.size();i++)printf("%d ",A[i]);return 0;
}

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define Max(a,b) (a>b?a:b)
#define N 200005
using namespace std;
int read()
{int x=0,f=1;char c=getchar();while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}return x*f;
}
int n,m,head[N],cnt=0,father1[N],dep[N],p[N][20],mlen[N][20];
bool used[N];
struct Node1
{int u,v,w;
}Edges1[N];
bool operator < (Node1 A,Node1 B){return A.w<B.w;}
struct Node2
{int nxt,to,w;
}Edges2[N*2];
int findf(int x)
{return (father1[x]==x)?x:(father1[x]=findf(father1[x]));}
void addedge(int u,int v,int w)
{Edges2[++cnt].nxt=head[u];head[u]=cnt;Edges2[cnt].to=v,Edges2[cnt].w=w;
}
void dfs(int u)
{for(int i=head[u];~i;i=Edges2[i].nxt){int v=Edges2[i].to;if(v==p[u][0])continue;p[v][0]=u,dep[v]=dep[u]+1,mlen[v][0]=Edges2[i].w;dfs(v);}
}
void init()
{for(int i=1;(1<<i)<=n;i++)for(int j=1;j<=n;j++)p[j][i]=p[p[j][i-1]][i-1],mlen[j][i]=max(mlen[j][i-1],mlen[p[j][i-1]][i-1]);
}
int lca(int x,int y)
{int res=0;if(dep[x]>dep[y])swap(x,y);int f=dep[y]-dep[x];for(int i=0;(1<<i)<=f;i++)if(f&(1<<i))res=Max(res,mlen[y][i]),y=p[y][i];if(x!=y){for(int i=(int)log2(n);i>=0;i--)if(p[x][i]!=p[y][i])res=Max(res,mlen[x][i]),res=Max(res,mlen[y][i]),x=p[x][i],y=p[y][i];res=Max(res,mlen[x][0]),res=Max(res,mlen[y][0]),x=p[x][0];}return res;
}
int main()
{memset(head,-1,sizeof(head));memset(dep,0,sizeof(dep));memset(used,0,sizeof(used));n=read(),m=read();for(int i=1;i<=m;i++)Edges1[i].u=read(),Edges1[i].v=read(),Edges1[i].w=read();sort(Edges1+1,Edges1+1+m);for(int i=1;i<=n;i++)father1[i]=i;int cnt=0;for(int i=1;i<=m;i++){int fu=findf(Edges1[i].u),fv=findf(Edges1[i].v);if(fu!=fv){cnt++;addedge(Edges1[i].u,Edges1[i].v,Edges1[i].w);addedge(Edges1[i].v,Edges1[i].u,Edges1[i].w);used[i]=1;father1[fu]=fv; }if(cnt==n-1)break;}dep[1]=mlen[1][0]=0;p[1][0]=1;dfs(1),init();int ans=0;for(int i=1;i<=m;i++){if(used[i])continue;int u=Edges1[i].u,v=Edges1[i].v;int maxlen=lca(u,v);if(maxlen==Edges1[i].w)ans++;}printf("%d\n",ans);return 0;
} 

这是那个Second Solution…只需要在Kruskal时把权值相同的边批量处理。

权值相同的边如果在最开始也加不进去就永远不可能在MST中出现，可以加进去的边有able条，并查集合并实际能用到的边uni条，则其余able-uni条边都需要权值++使MST唯一。

（如果一条有可能加入MST的边由于其他边被先考虑而没加进去，那这条边端点u,v间路径上的最大边权与它的权值相同，即可以拆掉另一条换这条。）

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define Max(a,b) (a>b?a:b)
#define N 200005
using namespace std;
int read()
{int x=0,f=1;char c=getchar();while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}return x*f;
}
int n,m,head[N],cnt=0,father1[N],ans=0;
struct Node1
{int u,v,w;
}Edges1[N];
bool operator < (Node1 A,Node1 B){return A.w<B.w;}
int findf(int x)
{return (father1[x]==x)?x:(father1[x]=findf(father1[x]));}
int main()
{memset(head,-1,sizeof(head));n=read(),m=read();for(int i=1;i<=m;i++)Edges1[i].u=read(),Edges1[i].v=read(),Edges1[i].w=read();sort(Edges1+1,Edges1+1+m);for(int i=1;i<=n;i++)father1[i]=i;int cnt=0,able=0,uni=0,from=1;for(int i=1;i<=m;i++){int fu=findf(Edges1[i].u),fv=findf(Edges1[i].v);if(fu!=fv)able++;if(i==m||Edges1[i+1].w!=Edges1[i].w){for(int j=from;j<=i;j++){int fu=findf(Edges1[j].u),fv=findf(Edges1[j].v);if(fu!=fv)father1[fu]=fv,cnt++,uni++;}ans+=able-uni;able=uni=0,from=i;}}printf("%d\n",ans);return 0;
}