description
analysis
-
无向图上任意两点最大流不超过(2)，说明该图是仙人掌
-
由于最大流等于最小割，如果一条边在两个环里，断掉两个端点至少需要(3)的代价
-
对于该仙人掌，求两点间的最小割，那么要么割一条桥边，要么割一个环上的两条边
-
环上边权最小边一定要割，那就把这条边断开，环上其他边边权加上该边边权，相当于提前割
-
然后图就变成了一棵树，任意两点间最小割即为路径最小值
-
把(n-1)条树边从大到小插入，这样前面的不会影响到后面的贡献
-
并查集维护每一个集合的(sum p^{(i-1)n},sum p^i)，由于分配律所以两个集合的贡献相乘就是整个集合的贡献
-
我一开始(naive)认为要用(tarjan)缩环，打了才发现我还(too) (young)
-
(m)条边降序排序后先直接构树，记录构成环的边，然后找这些边端点(LCA)，暴力把路径上的边权加上
-
注意不要太过暴力，身经百战后倍增也是很重要的
code
#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 300005
#define MAXM 500005
#define ha 998244353
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)
#define rep(i,a) for (reg i=last[a];i;i=next[i])
using namespace std;
ll last[MAXM*2],next[MAXM*2],tov[MAXM*2],len[MAXM*2];
ll fa[MAXN],falen[MAXN],depth[MAXN];
ll sumin[MAXN],sumi[MAXN],anc[MAXN][20];
ll n,m,p,tot,ans,cnt;
bool bz[MAXN];
struct edge
{
ll x,y,z;
}a[MAXM],b[MAXM];
inline ll read()
{
ll x=0,f=1;char ch=getchar();
while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
return x*f;
}
inline ll max(ll x,ll y){return x>y?x:y;}
inline ll min(ll x,ll y){return x<y?x:y;}
inline bool cmp(edge a,edge b){return a.z>b.z;}
inline ll getfa(ll x){return fa[x]==x?x:fa[x]=getfa(fa[x]);}
inline ll pow(ll x,ll y)
{
ll z=1;
while (y){if (y&1)z=z*x%ha;x=x*x%ha,y>>=1;}
return z;
}
inline ll lca(ll x,ll y)
{
if (depth[x]<depth[y])swap(x,y);
fd(i,19,0)if (depth[anc[x][i]]>=depth[y])x=anc[x][i];
if (x==y)return x;
fd(i,19,0)if (anc[x][i]!=anc[y][i])x=anc[x][i],y=anc[y][i];
return anc[x][0];
}
inline void link(ll x,ll y,ll z){next[++tot]=last[x],last[x]=tot,tov[tot]=y,len[tot]=z;}
inline void dfs(ll x,ll y)
{
fo(i,1,19)anc[x][i]=anc[anc[x][i-1]][i-1];
rep(i,x)if (tov[i]!=y)fa[tov[i]]=anc[tov[i]][0]=x,falen[tov[i]]=len[i],depth[tov[i]]=depth[x]+1,dfs(tov[i],x);
}
int main()
{
freopen("T2.in","r",stdin);
//freopen("sakura.in","r",stdin);
//freopen("sakura.out","w",stdout);
n=read(),m=read(),p=read();
fo(i,1,n)fa[i]=i;
fo(i,1,m)a[i].x=read(),a[i].y=read(),a[i].z=read();
sort(a+1,a+m+1,cmp);
fo(i,1,m)
{
ll x=a[i].x,y=a[i].y,z=a[i].z,xx=getfa(x),yy=getfa(y);
if (xx!=yy)fa[xx]=yy,link(x,y,z),link(y,x,z);
else b[++cnt].x=x,b[cnt].y=y,b[cnt].z=z;
}
depth[1]=1,dfs(1,0),tot=0;
fo(i,1,cnt)
{
ll x=b[i].x,y=b[i].y,z=b[i].z,LCA=lca(x,y);
if (LCA!=x){falen[x]+=z;while (fa[x]!=LCA)x=fa[x],falen[x]+=z;}
if (LCA!=y){falen[y]+=z;while (fa[y]!=LCA)y=fa[y],falen[y]+=z;}
}
fo(i,2,n)a[++tot].x=i,a[tot].y=fa[i],a[tot].z=falen[i];
sort(a+1,a+n,cmp);
fo(i,1,n)sumin[i]=pow(p,(i-1)*n),sumi[i]=pow(p,i),fa[i]=i;
fo(i,1,n-1)
{
ll x=a[i].x,y=a[i].y,z=a[i].z,xx=getfa(x),yy=getfa(y);
(ans+=(z*sumin[xx]%ha*sumi[yy]%ha+z*sumin[yy]%ha*sumi[xx]%ha)%ha)%=ha;
fa[xx]=yy,(sumin[yy]+=sumin[xx])%=ha,(sumi[yy]+=sumi[xx])%=ha;
}
printf("%lld
",ans);
return 0;
}