Kruskal重构树裸题,

Sunshine互测的A题就是Kruskal重构树,我通过互测了解到了这个神奇的东西。。。

理解起来应该没什么难度吧,但是我的Peaks连WA,,,

省选估计要滚粗了TwT

#include<cstdio>
#include<cstring>
#include<algorithm>
#define for1(i,a,n) for(int i=(a);i<=(n);i++)
#define for2(i,a,n) for(int i=(a);i<(n);i++)
#define for3(i,a,n) for(int i=(a);i>=(n);i--)
#define for4(i,a,n) for(int i=(a);i>(n);i--)
#define read(x) x=getint()
#define CC(i,a) memset(i,a,sizeof(i))
using namespace std;
inline const int getint(){char c=getchar();int k=1,r=0;for(;c<'0'||c>'9';c=getchar())if(c=='-')k=-1;for(;c>='0'&&c<='9';c=getchar())r=r*10+c-'0';return k*r;}
inline const int max(const int &a,const int &b){return a>b?a:b;}
inline const int min(const int &a,const int &b){return a<b?a:b;}
inline void swapp(int &a,int &b){int c=a;a=b;b=c;}
const int N=15003;
const int M=30003;
struct node{int x,y,z;}E[M];
int n,m,lch[N<<1],rch[N<<1],num[N<<1],f[N<<1][16],fa[N<<1],deep[N<<1],cnt;
inline void init(){CC(lch,0);CC(rch,0);CC(num,0);CC(f,0);CC(fa,0);CC(deep,0);}
inline bool cmp(node X,node Y){return X.z<Y.z;}
inline int find(int X){
if (fa[X]==X) return X;
else {fa[X]=find(fa[X]); return fa[X];}
}
inline void LCA(){
for1(j,1,15)
for2(i,1,n<<1)
if (f[f[i][j-1]][j-1]!=0) f[i][j]=f[f[i][j-1]][j-1];
}
inline void dfs(int x){
if (lch[x]) {deep[lch[x]]=deep[x]+1; dfs(lch[x]);}
if (rch[x]) {deep[rch[x]]=deep[x]+1; dfs(rch[x]);}
}
inline int LCA_find(int u,int v){
if (deep[u]<deep[v]) swapp(u,v);
int dis=deep[u]-deep[v];
for1(i,0,15)
if ((1<<i)&dis) u=f[u][i];
if (u==v) return u;
for3(i,15,0) if (f[u][i]!=f[v][i]){u=f[u][i]; v=f[v][i];}
return f[u][0];
}
int main(){
init(); int K;
read(n); read(m); read(K);
for1(i,1,m) {read(E[i].x);read(E[i].y);read(E[i].z);}
sort(E+1,E+m+1,cmp);
cnt=n+1;
for2(i,1,n<<1) fa[i]=i;
for1(i,1,m){
int fx=find(E[i].x),fy=find(E[i].y);
if (fx==fy) continue;
fa[fx]=cnt; fa[fy]=cnt;
f[fx][0]=cnt; f[fy][0]=cnt;
lch[cnt]=fx; rch[cnt]=fy;
num[cnt]=E[i].z;
cnt++; if (cnt>(n<<1)-1) break;
}
LCA();
deep[(n<<1)-1]=1;
dfs((n<<1)-1);
int a,b,rt;
for1(i,1,K){
read(a); read(b);
rt=LCA_find(a,b);
printf("%d\n",num[rt]);
}return 0;
}

  然后就完了

04-30 19:10