大连网赛 1010 HDU5877 离散化+树状数组

题目题意还是很清楚，要统计二元组（u，v）的对数，要同时满足两个条件：①u是v的祖先，②u*v<=k。那么我可以在dfs的时候将离散化后的每个节点的值用树状数组更新，然后每到一个节点就用树状数组统计在祖先中有多少个小于k/a[now],记得在回溯的时候要消除兄弟的影响。用map实现离散化版本：#include<cstdio>#include<cstring>#include<vector>#include<map>#include<algorithm>using namespace std;#define LL long long#define lowbit(x) x&(-x)const int maxn=100010;int t,n,u,v,a[maxn],c[maxn],q[maxn],rd[maxn];vector<int> G[maxn];map<LL,int> mp;LL k,b[maxn];void update(int i,int x){while(i<=n){q[i]+=x;i+=lowbit(i);}}LL getsum(int i){LL res=0;while(i){res+=q[i];i-=lowbit(i);}return res;}int find(LL v){int l=1,r=n,mid;LL pos;while(l<=r){mid=(l+r)>>1;pos=b[mid];if(pos==v) return mid;else if(pos>v) r=mid-1;else l=mid+1;}if(b[l]>v) return l-1;else return l;}LL dfs(int now){LL ans=0,x=find(k/a[now]);ans+=getsum(x);update(c[now],1);int len=G[now].size();for(int i=0;i<len;i++) ans+=dfs(G[now][i]);update(c[now],-1);return ans;}int main(){scanf("%d",&t);while(t--){memset(q,0,sizeof q);mp.clear();scanf("%d%lld",&n,&k);for(int i=1;i<=n;i++){scanf("%d",&a[i]);G[i].clear();rd[i]=0;b[i]=(LL)a[i];}sort(b+1,b+n+1);for(int i=1;i<=n;i++) mp[b[i]]=i;b[n+1]=4e18;for(int i=1;i<=n;i++) c[i]=mp[a[i]];for(int i=1;i<n;i++){scanf("%d%d",&u,&v);G[u].push_back(v);rd[v]++;}int rt=-1;for(int i=1;i<=n;i++) if(!rd[i]) {rt=i;break;}printf("%lld\n",dfs(rt));}return 0;}自己写的离散化(竟然比上面的跑得慢=.=)#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>using namespace std;#define LL long long#define lowbit(x) x&(-x)const int maxn=100010;int t,n,u,v,a[maxn],c[maxn],rd[maxn];LL k,b[maxn];vector<int> G[maxn];void update(int i,int x){    while(i<=n)    {        c[i]+=x;        i+=lowbit(i);    }}LL getsum(int i){    LL res=0;    while(i)    {        res+=c[i];        i-=lowbit(i);    }    return res;}LL dfs(int now){    LL ans=0;    int len=G[now].size();    LL tmp1=upper_bound(b+1,b+n+1,(LL)k/a[now])-b;    LL tmp2=lower_bound(b+1,b+n+1,a[now])-b;    ans+=getsum(tmp1-1);    if(tmp2)update(tmp2,1);    for(int i=0;i<len;i++) ans+=dfs(G[now][i]);    if(tmp2)update(tmp2,-1);    return ans;}int main(){    scanf("%d",&t);    while(t--)    {        memset(c,0,sizeof c);        scanf("%d%lld",&n,&k);        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            b[i]=(LL)a[i];            rd[i]=0;            G[i].clear();        }        sort(b+1,b+n+1);        for(int i=1;i<n;i++)        {            scanf("%d%d",&u,&v);            G[u].push_back(v);            rd[v]++;        }        int rt=-1;        for(int i=1;i<=n;i++) if(!rd[i]){rt=i;break;}        printf("%lld\n",dfs(rt));    }    return 0;}