[NOIP 2016] 天天爱跑步

[题目链接]

           http://uoj.ac/problem/261

[算法]

          树上差分 , 时间复杂度 : O(N + M)

[代码]

       

#include<bits/stdc++.h>
using namespace std;
#define MAXN 1000010

struct edge
{
        int to,nxt;
} e[MAXN << 1];

int n,m,tot,timer;
bool visited[MAXN];
int f[MAXN],seq[MAXN],dfn[MAXN],fa[MAXN],dep[MAXN],w[MAXN],s[MAXN],t[MAXN],
        w1[MAXN],w2[MAXN],ans[MAXN],head[MAXN],size[MAXN],lca[MAXN],cnt1[MAXN],cnt2[MAXN];
vector< pair<int,int> > q[MAXN],tag1[MAXN],tag2[MAXN],pos[MAXN];

template <typename T> inline void read(T &x)
{
    int f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; 
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline void addedge(int u,int v)
{
        tot++;
        e[tot] = (edge){v,head[u]};
        head[u] = tot;
}
inline int get_root(int x)
{
        if (f[x] == x) return x;
        return f[x] = get_root(f[x]);
}
inline void tarjan(int u,int father)
{
        dfn[u] = ++timer;
        seq[timer] = u;
        size[u] = 1;
        visited[u] = true;
        for (int i = 0; i < (int)q[u].size(); i++)
        {
                int v = q[u][i].first , id = q[u][i].second;
                if (visited[v]) lca[id] = get_root(v); 
        }
        for (int i = head[u]; i; i = e[i].nxt)
        {
                int v = e[i].to;
                if (v == father) continue;
                dep[v] = dep[u] + 1;
                tarjan(v,u);
                fa[v] = f[v] = u;
                size[u] += size[v];
        }
}
int main() 
{
        
        read(n); read(m);
        for (int i = 1; i < n; i++)
        {
                int u,v;
                read(u); read(v);
                addedge(u,v);
                addedge(v,u);
        }
        for (int i = 1; i <= n; i++) read(w[i]);
        for (int i = 1; i <= n; i++) f[i] = i;
        for (int i = 1; i <= m; i++)
        {
                read(s[i]);
                read(t[i]);    
                q[s[i]].push_back(make_pair(t[i],i));
                q[t[i]].push_back(make_pair(s[i],i));
        } 
        tarjan(1,-1);
        for (int i = 1; i <= n; i++) w1[i] = dep[i] + w[i];
        for (int i = 1; i <= n; i++) w2[i] = dep[i] - w[i];
        for (int i = 1; i <= m; i++)
        {
                if (lca[i] == s[i]) 
                {
                        tag2[t[i]].push_back(make_pair(dep[s[i]],1));
                        tag2[fa[s[i]]].push_back(make_pair(dep[s[i]],-1));
                } else if (lca[i] == t[i])
                {
                        tag1[s[i]].push_back(make_pair(dep[s[i]],1));
                        tag1[fa[t[i]]].push_back(make_pair(dep[s[i]],-1));
                } else
                {
                        tag1[s[i]].push_back(make_pair(dep[s[i]],1));
                        tag1[fa[lca[i]]].push_back(make_pair(dep[s[i]],-1));
                        tag2[t[i]].push_back(make_pair(2 * dep[lca[i]] - dep[s[i]],1));
                        tag2[lca[i]].push_back(make_pair(2 * dep[lca[i]] - dep[s[i]],-1));
                }
        }
        for (int i = 1; i <= n; i++)
        {
                pos[dfn[i] - 1].push_back(make_pair(i,-1));
                pos[dfn[i] + size[i] - 1].push_back(make_pair(i,1));
        }
        int val = 2 * n;
        for (int i = 1; i <= n; i++)
        {
                int u = seq[i];
                for (int j = 0; j < (int)tag1[u].size(); j++)
                        cnt1[tag1[u][j].first + val] += tag1[u][j].second;
                for (int j = 0; j < (int)tag2[u].size(); j++)
                        cnt2[tag2[u][j].first + val] += tag2[u][j].second; 
                for (int j = 0; j < (int)pos[i].size(); j++)
                {
                        ans[pos[i][j].first] += cnt1[w1[pos[i][j].first] + val] * pos[i][j].second;
                        ans[pos[i][j].first] += cnt2[w2[pos[i][j].first] + val] * pos[i][j].second; 
                }
        }
        for (int i = 1; i < n; i++) printf("%d ",ans[i]);
        printf("%d\n",ans[n]);
        
        return 0;
    
}