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: >>>>>>
", (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 + 100;
LL vv[2][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*2];
int fi[maxn], nxt[maxn*2];
LL q[2][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+1, v, -val);
}

void query(int x, LL &res1, LL &res2)
{
    res1 = 0, res2 = 0;
    while(x > 0)
    {
        res1 += vv[0][x];
        res2 += vv[1][x];
        x -= lowbit(x);
    }
}
void dfs1(int u, int f)
{
    fa[u] = f;
    dep[u] = dep[f] + 1;
    sz[u] = 1;
    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 = 0; i <= n; i++)
    {
        vv[0][i] = vv[1][i] = 0;
        son[i] = 0;
        sz[i] = 0;
        fa[i] = 0;
        cnt = 0;
        fi[i] = 0;
        nxt[i<<1] = nxt[i<<1|1] = 0;
        q[0][i] = q[1][i] = 0;
    }
}
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 = 1; i < n; i++)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        add(u, v, i);
    }
    cnt = 0;
}
inline bool isdigit(char ch)
{
    return ch >= '0' && ch <= '9';
}
const LL B = (int)1e5 + 10;

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[0][num[top[u]]] += k;
        q[0][num[u]+1] -= k;
        u = fa[top[u]];
    }
    if(dep[u] < dep[v]) swap(u, v);
    q[0][num[v]] += k;
    q[0][num[u]+1] -= 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[1][num[son[top[u]]]] += k;
            q[1][num[u]+1] -= k;
        }
        u = top[u];
        q[1][num[u]] += k;
        q[1][num[u]+1] -= k;;
        u = fa[u];
    }
    if(dep[u] < dep[v]) swap(u, v);
    if(u != v)
    {
        q[1][num[son[v]]] += k;
        q[1][num[u]+1] -=  k;
    }
}
LL ans[maxn];

void solve()
{
    dfs1(1, 0);
    top[1] = 1;
    num[1] = ++cnt;
    dfs2(1, 0);
    for(int i = 0; i < m; i++)
    {
        char s[20];
        int u, v, k;
        scanf("%s%d%d%d", s, &u, &v, &k);
        if(strcmp(s, "ADD1") == 0)
        {
            update(u, v, k);
        }
        else
        {
            update1(u, v, k);
        }
    }
    for(int k = 0; k < 2; k++){
            for(int i = 1; i <= n; i++)
                update(i, vv[k], q[k][i]);
        }

    for(int i = 1; i <= n; i++)
    {
        LL res1, res2;
        query(num[i], res1, res2);
        ans[num[i]] = res2;
        printf("%I64d%c", res1, i == n ? '
' : ' ');
    }
    for(int i = 1; i <= n-1; i++)
    {
        LL res = ans[ID[i]];
        printf("%I64d%c", res, i == n-1 ? '
' : ' ');
    }
    if(n == 1)
        puts("");
}
int main()
{

    int T;
    for(int t = scanf("%d", &T); t <= T; t++)
    {
        printf("Case #%d:
", t);
        input();
        solve();
    }
    return 0;
}