HDU5029 Relief grain 树链剖分+差分统计答案

大致题意：给出一棵n个节点有根树，现在给m个x、y，使得x到y路径上所有点加上标记z，现需要统计每个节点中数量最多的标记种类

先考虑线性序列，在x-y添加标记z，利用差分思想，在x处添加z，在y+1减去z，然后用一个维护标记数的线段树顺序维护，每个节点询问数量最多的节点即可。然后树型结构转线性，利用树剖即可。

#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <cstdlib>#include <algorithm>#include <iostream>#include <vector>using namespace std;const int maxn = 100010;struct edge{    int v,next;}e[maxn << 1];int h[maxn],num;struct node{    int l,r,x,y;}t[maxn << 1];int tot,cnt,ans[maxn];int fa[maxn],son[maxn],pos[maxn],ppos[maxn],top[maxn],size[maxn],dep[maxn];vector<int>a[maxn];int n,q;void add_edge(int u,int v){    num++;    e[num].v = v;    e[num].next = h[u];    h[u] = num;}int u,v;void build(int p,int l,int r){    int mid = l + r >> 1;    if(l == r)    {        t[p].x = l;        t[p].y = 0;        return;    }    t[p].x = 0;    t[p].y = 0;    t[p].l = ++tot;    t[p].r = ++tot;    build(t[p].l,l,mid);    build(t[p].r,mid+1,r);}void change(int p,int l,int r,int a,int b){    int mid = l + r >> 1;    if(l == r)    {        t[p].y += b;        return;    }    if(a <= mid)        change(t[p].l,l,mid,a,b);    else if(a > mid)        change(t[p].r,mid+1,r,a,b);    if(t[t[p].l].y >= t[t[p].r].y)    {        t[p].x = t[t[p].l].x;        t[p].y = t[t[p].l].y;    }    else    {        t[p].x = t[t[p].r].x;        t[p].y = t[t[p].r].y;    }}void dfs1(int x){    dep[x] = dep[fa[x]] + 1;    size[x] = 1;    son[x] = 0;    for(int i = h[x]; i; i = e[i].next)    {        if(e[i].v == fa[x])            continue;        fa[e[i].v] = x;        dfs1(e[i].v);        if(size[e[i].v] > size[son[x]])            son[x] = e[i].v;        size[x] += size[e[i].v];    }}void dfs2(int x){    if(son[fa[x]] == x)        top[x] = top[fa[x]];    else    {        top[x] = x;        for(int i = x; i; i = son[i])        {            pos[i] = ++cnt;            ppos[cnt] = i;        }    }    for(int i = h[x]; i; i = e[i].next)        if(e[i].v != fa[x])            dfs2(e[i].v);}void updata(int x,int y,int z){    int fx = top[x];    int fy = top[y];    while(fx != fy)    {        if(dep[fx] > dep[fy])        {            a[pos[x]+1].push_back(-z);            a[pos[top[x]]].push_back(z);            x = fa[fx];            fx = top[x];        }        else        {            a[pos[y]+1].push_back(-z);            a[pos[top[y]]].push_back(z);            y = fa[fy];            fy = top[y];        }    }    if(pos[x] > pos[y])    {        a[pos[y]].push_back(z);        a[pos[x]+1].push_back(-z);    }    else    {        a[pos[x]].push_back(z);        a[pos[y]+1].push_back(-z);    }}int main(){    while(scanf("%d%d",&n,&q), n != 0 || q != 0)    {        num = cnt = tot = 0;        memset(h,0,sizeof(h));        memset(fa,0,sizeof(fa));        memset(size,0,sizeof(size));        memset(son,0,sizeof(size));        memset(top,0,sizeof(top));        memset(dep,0,sizeof(dep));        for(int i = 1; i <= n; i++)            a[i].clear();        for(int i = 0; i < n - 1; i++)        {            scanf("%d%d",&u,&v);            add_edge(u,v);            add_edge(v,u);        }        dfs1(1);        dfs2(1);        int x,y,z;        int nn = 0;        while(q--)        {            scanf("%d%d%d",&x,&y,&z);                updata(x,y,z);            nn = max(nn,z);        }        build(0,0,nn);        for(int i = 1; i <= n; i++)        {            for(int j = 0; j < a[i].size(); j++)            {                if(a[i][j] > 0)                    change(0,0,nn,a[i][j],1);                else                    change(0,0,nn,-a[i][j],-1);                }            ans[ppos[i]] = t[0].x;        }        for(int i = 1; i <= n; i++)            printf("%d\n",ans[i]);    }    return 0;}