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哈夫曼树。
如果要最大的深度最小,再按h排序即可。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<queue> #define maxv 100500 #define maxe 1000500 using namespace std; long long n,k,val[maxv*5],nume=0,g[maxv*5],tot,sum=0,dis[maxv*5],fath[maxv*5]; struct edge { long long v,nxt; }e[maxe]; struct status { long long val,pnt,h; }; bool operator <(status x,status y) { if (x.val!=y.val) return x.val>y.val; return x.h>y.h; } priority_queue <status> q; void addedge(long long u,long long v) { e[++nume].v=v; e[nume].nxt=g[u]; g[u]=nume; } void dfs(long long x) { for (long long i=g[x];i;i=e[i].nxt) { long long v=e[i].v; if (v!=fath[x]) { fath[v]=x;dis[v]=dis[x]+1; dfs(v); } } } int max(int a,int b) { if (a>b) return a; return b; } int main() { scanf("%lld%lld",&n,&k); for (long long i=1;i<=n;i++) { scanf("%lld",&val[i]); status s; s.pnt=i;s.val=val[i];s.h=0; q.push(s); } long long ret=0; while ((n+ret-1)%(k-1)!=0) ret++; for (long long i=1;i<=ret;i++) { status s; s.pnt=n+i;s.val=0;s.h=0; q.push(s); val[n+i]=0; } n+=ret;tot=n; long long ret1=0,ret2=0; while (q.size()>1) { sum=0;tot++;int mx=0; for (long long i=1;i<=k;i++) { status b=q.top();q.pop(); sum+=b.val;mx=max(mx,b.h); addedge(b.pnt,tot);addedge(tot,b.pnt); } status s; s.pnt=tot;s.val=sum;s.h=mx+1;q.push(s); ret2=max(ret2,s.h); val[tot]=0; } dfs(tot); for (long long i=1;i<=n;i++) ret1+=val[i]*dis[i]; printf("%lld\n%lld\n",ret1,ret2); return 0; }
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原文地址:http://www.cnblogs.com/ziliuziliu/p/5843456.html