BZOJ 1036 树的统计 树链剖分or link-cut-tree

    崩溃，从上个星期开始BZOJ就一直RE，死命RE，写什么都能RE，心好累。还好吧！算了，不管它。幸亏有师兄，把那个测试数据给我测了一下，否则我都快抑郁了。加油吧！

    这道题就是一个模版题了，不过也学了点东西，比如写线段树时，数组要开到4倍。

    接下来没办法了，只能是我RE的代码了。（如果有大神知道哪里RE了，麻烦告知一声）。

#include<cstdio>
#include<iostream>
#include<vector>
#define rep(i,j,k) for(int i = j; i <= k; i++)
#define maxn 300500
#define INF 0x7fffffff
using namespace std;
  
int n, key[maxn] = {0};
vector<int> q[maxn];
  
int pa[maxn] = {0}, son[maxn] = {0}, size[maxn] = {0}, top[maxn] = {0};
int dep[maxn] = {0}, w[maxn] = {0}, tot = 0, maxl = -INF;
  
void dfs(int now)
{
    son[now] = 0;
    size[now] = 1;
    int s = q[now].size();
    rep(i,0,s-1)
    {
        int to = q[now][i];
        if( to == pa[now] ) continue;
        pa[to] = now;
        dep[to] = dep[now] + 1;
        dfs(to);
        size[now] += size[to];
        if( size[to] > size[son[now]] ) son[now] = to;
    }
}
  
void dfs2(int now,int ph)
{
    top[now] = ph;
    w[now] = ++tot;
    if( son[now] ) dfs2(son[now],ph);
    int s = q[now].size();
    rep(i,0,s-1)
    {
        int y = q[now][i];
        if( y != son[now] && y != pa[now] ) dfs2(y,y);
    }
}
  
struct node{
int l, r, maxl, sum;
node(){
    l = r = maxl = sum = 0;
}
};
node nod[maxn*4];
  
void yuchu(int l,int r,int num)
{
    nod[num].l = l, nod[num].r = r;
    if( l == r ){
        nod[num].maxl = nod[num].sum = key[l];
        return;
    }
    int mid = (l+r) / 2;
    yuchu(l,mid,num*2);
    yuchu(mid+1,r,num*2+1);
    nod[num].maxl = max(nod[num*2].maxl,nod[num*2+1].maxl);
    nod[num].sum = nod[num*2].sum + nod[num*2+1].sum;
}
  
void update(int po,int key,int num)
{
    int l = nod[num].l, r = nod[num].r;
    if( l == r ){
        nod[num].sum = nod[num].maxl = key;
        return;
    }
    int mid = (l+r) / 2;
    if( po <= mid ){
        update(po,key,num*2);
    }
    else {
        update(po,key,num*2+1);
    }
    nod[num].maxl = max(nod[num*2].maxl,nod[num*2+1].maxl);
    nod[num].sum = nod[num*2].sum + nod[num*2+1].sum;
}
  
int qure(int ll,int rr,int num)
{
    int l = nod[num].l, r = nod[num].r;
    if( l>=ll && rr >= r ){
        if( nod[num].maxl > maxl ) maxl = nod[num].maxl;
        return nod[num].sum;
    }
    int mid = (l+r) / 2;
    if( rr <= mid ){
        return qure(ll,rr,num*2);
    }
    else if( ll > mid ){
        return qure(ll,rr,num*2+1);
    }
    else if( ll <= mid && rr > mid ){
        return qure(ll,mid,num*2) + qure(mid+1,rr,num*2+1);
    }
}
  
int quire(int x,int y,int pan)
{
    int ans = 0;
    maxl = -INF;
    while( top[x] != top[y] ){
        if( dep[top[x]] < dep[top[y]] ){
            swap(x,y);
        }
        ans += qure(w[top[x]],w[x],1);
        x = pa[top[x]];
    }
    if( dep[x] > dep[y] ) swap(x,y);
    ans += qure(w[x],w[y],1);
    if( pan ) return maxl;
    else return ans; 
}
  
int main()
{
    scanf("%d", &n);
    int x, y;
    rep(i,1,n-1){
        scanf("%d %d", &x, &y);
        q[x].push_back(y);
        q[y].push_back(x);
    }
    dfs(1);
    dfs2(1,1);
    rep(i,1,n){
        scanf("%d", &key[w[i]]);
    }
    yuchu(1,tot,1);
    int q, key;
    scanf("%d", &q); char c[7] = {0};
    rep(i,1,q){
        scanf("%s", c);
        if( c[0] == 'Q' ){
            if( c[1] == 'S' ){
                scanf("%d %d", &x, &y);
                cout<<quire(x,y,0)<<endl;
            }
            else if( c[1] == 'M' ){
                scanf("%d %d", &x, &y);
                cout<<quire(x,y,1)<<endl;
            }
        }
        else if( c[0] == 'C' ){
            scanf("%d %d", &x, &key);
            update(w[x],key,1);
        }
    }
    return 0;
}

link-cut-tree 版本这是我一直想学的数据结构，（这个坑已经挖了好久了，直到这个星期才开始填，还好填了大半，虽然有一些还不大理解，再多做两道题吧！）

加油，相信自己。四千多毫秒，比树链剖分慢多了。  注意一开始要把mx[0] 弄得很小，否则会出错。

#include<cstdio>
#include<iostream>
#define rep(i,j,k) for(int i = j; i <= k; i++)
#define maxn 30005
using namespace std;

int read()
{
    int s = 0, t = 1; char c = getchar();
    while( !isdigit(c) ){
        if( c =='-' ) t = -1; c = getchar();
    }
    while( isdigit(c) ){
        s = s * 10 + c - '0'; c=  getchar();
    }
    return s * t;
}

int mx[maxn], c[maxn][2], sum[maxn], w[maxn];
int u[maxn], v[maxn], q[maxn], fa[maxn], n, m, top;
bool rev[maxn];

bool root(int x)
{
    return c[fa[x]][0] != x && c[fa[x]][1] != x;
}

void pushdown(int x)
{
    if( rev[x] ){
        int l = c[x][0], r = c[x][1];
        rev[x] ^= 1, rev[l] ^= 1, rev[r] ^= 1; 
        swap(c[x][0],c[x][1]);
    }
}

void update(int x)
{
    int l = c[x][0], r = c[x][1];
    sum[x] = sum[l] + sum[r] + w[x];
    mx[x] = max(max(mx[l],mx[r]),w[x]);
} 

void rorate(int x)
{
    int y = fa[x], z = fa[y];
    int l = (c[y][0] == x), r = l ^ 1; 
    if( !root(y) ) c[z][y==c[z][1]] = x;
    fa[y] = x, fa[x] = z; if( c[x][l] ) fa[c[x][l]] = y;
    c[y][r] = c[x][l]; c[x][l] = y; 
    update(y), update(x);
}

void splay(int x)
{
    q[++top] = x;
    for(int i = x; !root(i); i=fa[i]){
        q[++top] = fa[i];
    }
    while( top ) pushdown(q[top--]);
    while( !root(x) ){
        int y = fa[x], z = fa[y];
        if( !root(y) ){
            if( c[y][0] == x ^ c[z][0] == y ) rorate(x);
            else rorate(y);
        }
        rorate(x);
    }
}

void access(int x)
{
    for(int t = 0; x; t = x, x = fa[x]){
        splay(x), c[x][1] = t, update(x);
    }
}

void makeroot(int x)
{
    access(x), splay(x); rev[x] ^= 1;
}

void join(int x, int y)
{
    makeroot(x), fa[x] = y; 
}

void split(int x, int y)
{
    makeroot(x), access(y), splay(y);
}

int main()
{
    n = read(), mx[0] = -0x7fffffff;
    rep(i,1,n-1){
        u[i] = read(), v[i] = read();
    }
    rep(i,1,n){
        sum[i] = w[i] = mx[i] = read();
    }
    rep(i,1,n-1){
        join(u[i],v[i]);
    }
    m = read(); char ch[10];
    rep(i,1,m){
        scanf("%s", ch);
        int u = read(), v = read();
        if( ch[1] == 'H' ){
            splay(u);
            w[u] = v;
            update(u);
        }
        if( ch[1] == 'M' ){
            split(u,v); printf("%d
", mx[v]);
        }
        if( ch[1] == 'S' ) {
            split(u,v); printf("%d
", sum[v]);
        }
    }
    return 0;
}