HDU 5044 TREE

题意：一棵树上两种操作，操作1，改变u到v的每一点的值增加k，操作2，改变u到v每一条边值增加k。最后结束时问，每一点和每一条边的值。

初始时，点和边的值都为0.

分析：

很显然要用树链剖分，将点和边分别划分成连续一段的编号，然后就是维护一段一段的值了，给它增加一个值，由于题目只需要输出最后结果，那么可以用树桩数组维护。

以边为例，对于l~r的每个值增加k，利用树桩数组v[i]表示第i个值与前一个值的差值，那么第k条边的值就是sum{v[i]| 1 <= i <= k},更新时则只要给v[l]加k，给v[r+1]加(- k)， 这个都可以累加下来，最后依次对每个v[i]更新，然后求出结果就行了。

代码：

 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <map>
 #include <cstring>
 #include <set>
 #include <bitset>
 #include <cstdio>
 #include <cmath>
 #define esp 1e-8
 #pragma comment(linker, "/STACK:102400000,102400000")
 #define in freopen("F:\\rootial\\data\\data.txt", "r", stdin);
 #define IN freopen("solve_in.txt", "r", stdin);
 #define out freopen("out.txt", "w", stdout);
 #define pf(x) ((x)*(x))
 #define lowbit(x) ((x)&(-(x)))
 #define bug(x) printf("Line %d: >>>>>>\n", (x));
 #define lson l, m, rt<<1
 #define rson m+1, r, rt<<1|1
 #define pb push_back
 #define mp make_pair
 #define pi acos(-1.0)
 #define pf(x) ((x)*(x))

 using namespace std;
 typedef long long LL;
 typedef pair<int, int> PII;
 typedef map<LL,  LL> MPS;
 typedef MPS::iterator IT;

 const int maxn = (int)1e5 + ;
 LL vv[][maxn];
 int n, m;
 int top[maxn], num[maxn], tnum[maxn], ID[maxn], son[maxn], sz[maxn], fa[maxn], dep[maxn], tmp[maxn];
 int cnt;

 struct Edge
 {
     int v, id;
     Edge() {}
     Edge(int v, int id):v(v), id(id) {}
 } edge[maxn*];
 int fi[maxn], nxt[maxn*];
 LL q[][maxn];

 void update(int x, LL v[], LL val)
 {
     while(x <= n)
     {
         v[x] +=val;
         x += lowbit(x);
     }
 }
 void update(int l, int r, LL val, LL v[])
 {
     update(l, v, val);
     update(r+, v, -val);
 }

 void query(int x, LL &res1, LL &res2)
 {
     res1 = , res2 = ;
     while(x > )
     {
         res1 += vv[][x];
         res2 += vv[][x];
         x -= lowbit(x);
     }
 }
 void dfs1(int u, int f)
 {
     fa[u] = f;
     dep[u] = dep[f] + ;
     sz[u] = ;
     for(int i = fi[u]; i; i = nxt[i])
     {
         int v = edge[i].v;
         int id = edge[i].id;
         if(v == f)
             continue;
         tmp[v] = id;
         dfs1(v, u);
         sz[u] += sz[v];
         if(sz[son[u]] < sz[v])
             son[u] = v;
     }
 }
 void dfs2(int u, int f)
 {
     if(son[u])
     {
         num[son[u]] = ++cnt;
         top[son[u]] = top[u];
         ID[tmp[son[u]]] = cnt;
         dfs2(son[u], u);
     }
     for(int i = fi[u]; i; i = nxt[i])
     {
         int v = edge[i].v;
         int id = edge[i].id;
         if(v == f || v == son[u]) continue;
         top[v] = v;
         num[v] = ++cnt;
         ID[tmp[v]] = cnt;
         dfs2(v, u);
     }
 }
 void init()
 {
     for(int i = ; i <= n; i++)
     {
         vv[][i] = vv[][i] = ;
         son[i] = ;
         sz[i] = ;
         fa[i] = ;
         cnt = ;
         fi[i] = ;
         nxt[i<<] = nxt[i<<|] = ;
         q[][i] = q[][i] = ;
     }
 }
 void add(int u, int v, int x)
 {
     edge[++cnt] = Edge(v, x);
     nxt[cnt] = fi[u];
     fi[u] = cnt;
     edge[++cnt] = Edge(u, x);
     nxt[cnt] = fi[v];
     fi[v] = cnt;
 }
 void input()
 {
     scanf("%d%d", &n, &m);
     init();
     for(int i = ; i < n; i++)
     {
         int u, v;
         scanf("%d%d", &u, &v);
         add(u, v, i);
     }
     cnt = ;
 }
 inline bool isdigit(char ch)
 {
     return ch >= '' && ch <= '';
 }
 const LL B = (int)1e5 + ;

 void update(int u, int v, int k) //dian
 {
     while(top[u] != top[v])
     {
         if(dep[top[u]] < dep[top[v]]) swap(u, v);
         q[][num[top[u]]] += k;
         q[][num[u]+] -= k;
         u = fa[top[u]];
     }
     if(dep[u] < dep[v]) swap(u, v);
     q[][num[v]] += k;
     q[][num[u]+] -= k;
 }
 void update1(int u, int v, int k) //bian
 {
     while(top[u] != top[v])
     {
         if(dep[top[u]] < dep[top[v]]) swap(u, v);
         if(u != top[u])
         {
             q[][num[son[top[u]]]] += k;
             q[][num[u]+] -= k;
         }
         u = top[u];
         q[][num[u]] += k;
         q[][num[u]+] -= k;;
         u = fa[u];
     }
     if(dep[u] < dep[v]) swap(u, v);
     if(u != v)
     {
         q[][num[son[v]]] += k;
         q[][num[u]+] -=  k;
     }
 }
 LL ans[maxn];

 void solve()
 {
     dfs1(, );
     top[] = ;
     num[] = ++cnt;
     dfs2(, );
     for(int i = ; i < m; i++)
     {
         char s[];
         int u, v, k;
         scanf("%s%d%d%d", s, &u, &v, &k);
         if(strcmp(s, "ADD1") == )
         {
             update(u, v, k);
         }
         else
         {
             update1(u, v, k);
         }
     }
     for(int k = ; k < ; k++){
             for(int i = ; i <= n; i++)
                 update(i, vv[k], q[k][i]);
         }

     for(int i = ; i <= n; i++)
     {
         LL res1, res2;
         query(num[i], res1, res2);
         ans[num[i]] = res2;
         printf("%I64d%c", res1, i == n ? '\n' : ' ');
     }
     for(int i = ; i <= n-; i++)
     {
         LL res = ans[ID[i]];
         printf("%I64d%c", res, i == n- ? '\n' : ' ');
     }
     if(n == )
         puts("");
 }
 int main()
 {

     int T;
     for(int t = scanf("%d", &T); t <= T; t++)
     {
         printf("Case #%d:\n", t);
         input();
         solve();
     }
     return ;
 }