[poi2007]mem

题意：给定点数n<=300000的一棵树，然后初始时每条边权值为1，接下来按照时间点先后顺序的n+m-1个操作，

操作有两种：

1.A a b 把a到b的边权改为0

2.W u 求1号点到u号点的路径权值和

思路：如果现在把题目简化为是在一条直线上的操作，每次在中间删除相邻边权值或者查询某个点到1号点的权值和

我们很容易想把询问离线，然后从1->n扫描1遍，然后用一个树状数组维护前缀和即可。。

到了本题利用dfs序显然就可以转化成线性模型，

具体的话

做到点u，

如果有一个操作1在(u, fa[u])的边，时间为t，那么在t时间点删除一个点

如果有一个操作2在u点，时间为t，那么就等价于查询1~u路径上的点数-t时间内删除的点数

回溯时把操作还原

code(stl)：

 #pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<map>
 #include<set>
 #include<vector>
 #include<queue>
 #include<stack>
 #include<ctime>
 #define x first
 #define y second
 #define M0(x)  memset(x, 0, sizeof(x))
 #define vii vector<int>::iterator
 #define vpi vector<pair<int, int> >::iterator
 #define Inf  0x7fffffff
 using namespace std;
 typedef pair<int, int> pii;
 const int maxn = ;
 vector<int> e[maxn];
 vector<pii> q[maxn];
 int n, m, ans[maxn<<];
 int pos[maxn<<];

 inline void R(int &ret){
     ret = ;
     bool ok = ;
     for( ; ;){
         int c = getchar();
         if (c >= '' && c <= '') ret = (ret << ) + (ret << ) + c - '', ok = ;
         else if (ok) return;
     }
 }

 void init(){
      int u, v;
     // for (int i = 1; i <= n; ++i) e[i].clear(), q[i].clear();
      pii tmp;
      for (int i = ; i < n; ++i)
           R(u), R(v), e[u].push_back(v), e[v].push_back(u);
      char op[];
      R(m);
      int nm = n + m - ;
      int n1 = ;
      for (int i = ; i <= nm; ++i){
            scanf("%s", op);
            if (op[] == 'A') ++n1;
            tmp.x = i;
            pos[i] = n1;
            if (op[] == 'W')
                   R(u), tmp.y = -, q[u].push_back(tmp);
            else {
                   R(u), R(v);
                   tmp.y = v, q[u].push_back(tmp);
                   tmp.y = u, q[v].push_back(tmp);
            }
      }
 //     cout << n1 << endl;
 }

 int vis[maxn], s[maxn], dep[maxn];
 void update(int x,const int& v){
      for (; x<=n; x += x&(-x)) s[x] += v;
 }

 int query(int x){
     int res = ;
     for (; x>; x -= x&(-x)) res += s[x];
     return  res;
 }

 int ss;
 void  dfs(const int& u){
       vis[u] = ;
       for (vpi it = q[u].begin(); it != q[u].end(); ++it)
           if (it->y != - && vis[it->y])  update(pos[it->x], );
       for (vpi it = q[u].begin(); it != q[u].end(); ++it)
           if (it->y == -)  ans[it->x] = dep[u] - query(pos[it->x]);
       for (vii it = e[u].begin(); it != e[u].end(); ++it)
           if (!vis[*it])  dep[*it] = dep[u] + , dfs(*it);
       for (vpi it = q[u].begin(); it != q[u].end(); ++it)
           if (it->y != - && dep[it->y] < dep[u])  update(pos[it->x], -);
 }

 void solve(){
      memset(ans, -, sizeof(int) * (n+m+));
      memset(vis, , sizeof(int) * (n+));
      memset(s, , sizeof(int) * (n+));
      dep[] = ;
      dfs();
      int nm = n + m;
      for (int i = ; i < nm; ++i)
           if (ans[i] >= ) printf("%d\n", ans[i]);
 }

 int main(){
     while (scanf("%d", &n) != EOF){
           init();
           solve();
     }
     return ;
 }

code（手写链表）：

 #pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<map>
 #include<set>
 #include<vector>
 #include<queue>
 #include<stack>
 #include<ctime>
 #define M0(x)  memset(x, 0, sizeof(x))
 #define Inf  0x7fffffff
 using namespace std;
 const int maxn = ;
 struct node{
        int v, next;
 } e[maxn * ];
 struct qes{
        int v, t, next;
 } q[maxn << ];
 int n, m, ans[maxn<<], last[maxn], lastq[maxn], tot, len;
 int pos[maxn<<];

 inline void R(int &ret){
     ret = ;
     bool ok = ;
     for( ; ;){
         int c = getchar();
         if (c >= '' && c <= '') ret = (ret << ) + (ret << ) + c - '', ok = ;
         else if (ok) return;
     }
 }

 inline void add(const int& u,const int& v){
     e[tot] = (node){v, last[u]}, last[u] = tot++;
 }

 inline void add_ask(const int& u,const int& v,const int& t){
     q[len] = (qes){v, t, lastq[u]}, lastq[u] = len++;
 }

 void init(){
      int u, v;
      memset(last, -, sizeof(last));
      memset(lastq, -, sizeof(lastq));
      len = tot = ;
      for (int i = ; i < n; ++i)
           R(u), R(v), add(u, v), add(v, u);
      char op[];
      R(m);
      int nm = n + m - ;
      int n1 = ;
      for (int i = ; i <= nm; ++i){
            scanf("%s", op);
            if (op[] == 'A') ++n1;
            pos[i] = n1;
            if (op[] == 'W')
                   R(u), add_ask(u, -, i);
            else {
                   R(u), R(v);
                   add_ask(u, v, i), add_ask(v, u, i);
            }
      }
 //     cout << n1 << endl;
 }

 int vis[maxn], s[maxn], dep[maxn];
 void update(int x,const int& v){
      for (; x<=n; x += x&(-x)) s[x] += v;
 }

 int query(int x){
     int res = ;
     for (; x>; x -= x&(-x)) res += s[x];
     return  res;
 }

 void  dfs(const int& u){
       vis[u] = ;
       for (int p = lastq[u]; ~p; p = q[p].next)
           if (q[p].v != - && vis[q[p].v])  update(pos[q[p].t], );
       for (int p = lastq[u]; ~p; p = q[p].next)
           if (q[p].v == -)  ans[q[p].t] = dep[u] - query(pos[q[p].t]);
       for (int p = last[u]; ~p; p = e[p].next)
           if (!vis[e[p].v])  dep[e[p].v] = dep[u] + , dfs(e[p].v);
       for (int p = lastq[u]; ~p; p = q[p].next)
           if (q[p].v != - && dep[q[p].v] < dep[u])  update(pos[q[p].t], -);
 }

 void solve(){
      memset(ans, -, sizeof(int) * (n+m+));
      memset(vis, , sizeof(int) * (n+));
      memset(s, , sizeof(int) * (n+));
      dep[] = ;
      dfs();
      int nm = n + m;
      for (int i = ; i < nm; ++i)
           if (ans[i] >= ) printf("%d\n", ans[i]);
 }

 int main(){
 //    freopen("a.in", "r", stdin);
 //    freopen("a.out", "w", stdout);
     while (scanf("%d", &n) != EOF){
           init();
           solve();
     }
     return ;
 }