詢問樹上距離為k的點對是否存在

直接n^2暴力處理點對 桶排記錄 可以過

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 1e5 + 5;
const int MAXM = 1e5 + 5;
int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], ed = 1;
int cost[MAXM << 1];
int ok[10000005
 
];
inline void addedge(int u, int v, int c) {
        to[++ed] = v;
        cost[ed] = c;
        nxt[ed] = Head[u];
        Head[u] = ed;
}
inline void ADD(int u, int v, int c) {
        addedge(u, v, c);
        addedge(v, u, c);
}
int n, m, k;
int sz[MAXN], f[MAXN], dep[MAXN], sumsz, root;
 
bool vis[MAXN];
int o[MAXN], cnt;
void getroot(int x, int fa) {
        sz[x] = 1;
        f[x] = 0;
        for (int i = Head[x]; i; i = nxt[i]) {
                int v = to[i];
                if (v == fa || vis[v]) {
                        continue;
                }
                getroot(v, x);
                sz[x] 
 
+= sz[v];
                f[x] = max(f[x], sz[v]);
        }
        f[x] = max(f[x], sumsz - sz[x]);
        if (f[x] < f[root]) {
                root = x;
        }
}
void getdeep(int x, int fa) {
        o[++cnt] = dep[x];
        for (int i = Head[x]; i; i = nxt[i]) {
                int v = to[i];
                if (v == fa || vis[v]) {
                        continue;
                }
                dep[v] = dep[x] + cost[i];
                getdeep(v, x);
        }
}
void calc(int x, int d, int add) {
        cnt = 0;
        dep[x] = d;
        getdeep(x, 0);
        sort(o + 1, o + cnt + 1);
        for (int i = 1; i <= cnt; i++) {
                for (int j = i + 1; j <= cnt; j++) {
                        ok[o[i] + o[j]] += add;
                }
        }
}
void solve(int x) {
        calc(x, 0, 1);
        vis[x] = 1;
        for (int i = Head[x]; i; i = nxt[i]) {
                int v = to[i];
                if (vis[v]) {
                        continue;
                }
                calc(v, cost[i], -1);
                root = 0, sumsz = sz[v];
                getroot(v, 0);
                solve(root);
        }
}
int main() {
        scanf("%d %d", &n, &m);
        cnt = 0;
        memset(Head, 0, sizeof(Head));
        memset(vis, 0, sizeof(vis));
        memset(ok, 0, sizeof(ok));
        ed = 1;
        int u, v, c;
        for (int i = 1; i < n; i++) {
                scanf("%d %d %d", &u, &v, &c);
                ADD(u, v, c);
        }
        root = 0, sumsz = f[0] = n;
        getroot(1, 0);
        solve(root);
        for (int i = 1; i <= m; i++) {
                scanf("%d", &k);
                if (ok[k]) {
                        printf("AYE\n");
                } else {
                        printf("NAY\n");
                }
        }
        return 0;
}

用類似poj1741的方法:對於每一次詢問 calc()函數中求出<=k的和>=k的數量再減去總對數 則為=k的數量

復雜度為O(m*n*log²n)

#include<bits/stdc++.h>
#define MAXN 10005
#define INF 1e9+7
using namespace std;
struct front_star{
    int to,next,w;
}edge[MAXN<<1];
int n,cnt=0,k,mx,root,ans=0,tot=1,siz,m;
int head[MAXN],sz[MAXN],temp[MAXN],idx[MAXN];
bool vis[MAXN];
int maxn(int a,int b)
{
    return a>b?a:b;
}
void addedge(int u,int v,int c)
{
    cnt++;
    edge[cnt].to=v;
    edge[cnt].w=c;
    edge[cnt].next=head[u];
    head[u]=cnt;
}
void findroot(int u,int fa)
{
    sz[u]=1;
    int msz=0;
    for(int i=head[u];~i;i=edge[i].next)
        {
            int v=edge[i].to;
            if(v!=fa&&!vis[v])
               {
                  findroot(v,u); 
                  sz[u]+=sz[v];
                  msz=maxn(msz,sz[v]);     
               }
        }
    msz=maxn(msz,siz-sz[u]);
    if(msz<mx)
       {
           mx=msz;
           root=u;
       }       
}
void init()
{
    memset(head,-1,sizeof(head));
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n-1;i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            addedge(a,b,c);
            addedge(b,a,c);
        }
}
void dist(int u,int fa)
{
    for(int i=head[u];~i;i=edge[i].next)
        {
            int v=edge[i].to;
            if(!vis[v]&&v!=fa)
               {
                   tot++;
                   idx[v]=tot;
                   temp[tot]=temp[idx[u]]+edge[i].w;
                   dist(v,u);
               }
        }
}
int count_ans(int u,int val)
{
    tot=1;
    idx[u]=1;
    temp[1]=val;
    dist(u,u);
    sort(temp+1,temp+1+tot);
    int L=1,R=tot,res1=0,res2=0,ret;
    while(L<=R)
          {
               if(temp[L]+temp[R]<=k)
                  {
                      res1+=R-L;
                      L++;
                  }
               else
                  R--;
          }
    L=1,R=tot;      
    while(L<=R)
          {
               if(temp[L]+temp[R]>=k)
                  {
                       res2+=R-L;
                       R--;
                  }
               else
                  L++;   
          } 
    ret=res1+res2-(tot*(tot-1))/2; 
    return ret;     
}
void divide(int u)
{
    ans+=count_ans(u,0);
    vis[u]=true;
    for(int i=head[u];~i;i=edge[i].next)
        {
            int v=edge[i].to;
            if(!vis[v]&&!vis[v])
               {
                    ans-=count_ans(v,edge[i].w);
                    siz=sz[v];
                    mx=INF;
                    findroot(v,u);
                    divide(root);  
               }
        }
}
void query()
{
    for(int i=1;i<=m;i++)
        {
            memset(vis,false,sizeof(vis));
            scanf("%d",&k);
            siz=n;
            mx=INF;
            ans=0;
            findroot(1,1);
            divide(root);
            if(ans==0)
               printf("NAY\n");
            else
               printf("AYE\n");   
        }
}
int main()
{
    init();
    query();
    return 0;
}

P3806 離線多次詢問 樹上距離為K的點對是否存在 點分治