bzoj3124(树形dp)

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求有多少条边：满足所有直径都经过该边。

两边树形dp，一遍以点做dp求出直径的个数，再以边做dp判断在直径上边的个数。

#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>using namespace std;const int N=200005;const int inf=0x3f3f3f3f;typedef long long ll;int n;int head[N],tot,ans;struct aa{int to,pre;ll dis;}edge[N*2];void addedge(int u,int v,ll d) {edge[++tot].to=v;edge[tot].pre=head[u];edge[tot].dis=d;head[u]=tot;}ll dis[N];ll MX,ANS,f[N];void dfs(int u,int fa){int v;ll tmp;dis[u]=0;f[u]=1;for (int i=head[u];i;i=edge[i].pre)if ((v=edge[i].to)!=fa){dfs(v,u);tmp=edge[i].dis+dis[v];if (tmp+dis[u]>MX) {MX=tmp+dis[u];ANS=f[v]*f[u];}else if (tmp+dis[u]==MX) ANS+=f[v]*f[u];if (tmp>dis[u]) dis[u]=tmp,f[u]=f[v];else if (tmp==dis[u]) f[u]+=f[v];}}void dfs(int u,int fa,ll mx,ll ff){ll tmp,fi=mx,se=0,fic=ff,sec=1;int v;for (int i=head[u];i;i=edge[i].pre)if ((v=edge[i].to)!=fa){tmp=dis[v]+edge[i].dis;if (tmp>fi) se=fi,sec=fic,fi=tmp,fic=f[v];else if (tmp==fi) fic+=f[v];else if (tmp>se) se=tmp,sec=f[v];else if (tmp==se) sec+=f[v];}for (int i=head[u];i;i=edge[i].pre) if ((v=edge[i].to)!=fa)if (dis[v]+edge[i].dis==fi){if (fic>f[v]) {if (dis[v]+edge[i].dis+fi==MX&&(fic-f[v])*f[v]==ANS) ans++;dfs(v,u,fi+edge[i].dis,fic-f[v]);} else {if (dis[v]+edge[i].dis+se==MX&&sec*f[v]==ANS) ans++;dfs(v,u,se+edge[i].dis,sec);}}else {if (dis[v]+edge[i].dis+fi==MX&&fic*f[v]==ANS) ans++; dfs(v,u,fi+edge[i].dis,fic);}}int main(){scanf("%d",&n);int u,v;ll d;for (int i=1;i<n;i++){scanf("%d%d%lld",&u,&v,&d);addedge(u,v,d);addedge(v,u,d);}dfs(1,0);dfs(1,0,0,1);//printf("%d\n",ANS);printf("%lld\n%d",MX,ans);return 0;}

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