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[模板]树链剖分

时间:2019-07-27 21:00:44      阅读:108      评论:0      收藏:0      [点我收藏+]

标签:cpp   ++   +=   ret   i++   head   mes   int   its   

#include<bits/stdc++.h>
#define lson(x) (x << 1)
#define rson(x) (x << 1 | 1)
using namespace std;

const int N = 100010,M = 200010;
int n,m,r,p;
long long a[N];
int d[N],son[N],fa[N],siz[N],top[N],rk[N],id[N],cnt;//树链剖分 
int head[N],edge[M],nxt[M],from[M],to[M],tot;//邻接表 
void add_edge(int x,int y){
    from[++tot] = x;
    to[tot] = y;
    nxt[tot] = head[x];
    head[x] = tot;
}

//线段树
struct Tree{
    int l,r;
    long long sum,tag;  
}t[N*4];
void push_up(int x) { t[x].sum = (t[lson(x)].sum + t[rson(x)].sum) % p; }
void push_down(int x){
    if(t[x].tag){
        t[lson(x)].tag = (t[lson(x)].tag + t[x].tag) % p; 
        t[lson(x)].sum = (t[lson(x)].sum + t[x].tag *(t[lson(x)].r - t[lson(x)].l + 1)) % p;
        t[rson(x)].tag = (t[rson(x)].tag + t[x].tag) % p; 
        t[rson(x)].sum = (t[rson(x)].sum + t[x].tag *(t[rson(x)].r - t[rson(x)].l + 1)) % p;
        t[x].tag = 0;
    }
}
void build(int x,int ll,int rr){
    t[x].l = ll; t[x].r = rr;
    if(ll == rr) { 
        t[x].sum = a[rk[ll]]; 
        if(t[x].sum > p) t[x].sum %= p; 
        return; 
    }
    int mid = (ll + rr) >> 1;
    build(lson(x),ll,mid);
    build(rson(x),mid + 1,rr);
    push_up(x);
}
void update(int x,int ll,int rr,int k){
    if(ll <= t[x].l && t[x].r <= rr){
        t[x].sum = (t[x].sum + k * (t[x].r - t[x].l + 1)) % p;
        t[x].tag = (t[x].tag + k) % p;
        return;
    }
    push_down(x);
    int mid = (t[x].l + t[x].r) >> 1;
    if(ll <= mid) update(lson(x),ll,rr,k);
    if(mid < rr)  update(rson(x),ll,rr,k);
    push_up(x);
}
long long query(int x,int ll,int rr){
    if(ll <= t[x].l && t[x].r <= rr) return t[x].sum;
    int mid = (t[x].l + t[x].r) >> 1;
    long long res = 0;
    push_down(x);
    if(ll <= mid) res = (res + query(lson(x),ll,rr)) % p;
    if(mid < rr)  res = (res + query(rson(x),ll,rr)) % p;
    push_up(x);
    return res;
}

//树链剖分 
void dfs1(int x){
    siz[x] = 1;
    for(int i = head[x]; i; i = nxt[i])
        if(!siz[to[i]]){
            d[to[i]] = d[x] + 1;
            fa[to[i]] = x;
            dfs1(to[i]);
            siz[x] +=  siz[to[i]];
            if(!son[x] || siz[to[i]] > siz[son[x]])
                son[x] = to[i];
        }
}
void dfs2(int x,int t){
    top[x] = t;
    id[x] = ++cnt;
    rk[cnt] = x;
    if(!son[x]) return;
    dfs2(son[x],t);
    for(int i = head[x]; i; i = nxt[i])
        if(to[i] != son[x] && to[i] != fa[x])
            dfs2(to[i],to[i]);
}
int main(){
//    freopen("testdata (6).in","r",stdin);
//    freopen("test.txt","w",stdout);
    cin >> n >> m >> r >> p;
    for(int i = 1; i <= n; i++)
        scanf("%lld",&a[i]);
    for(int i = 1; i < n; i++){
        int x,y;
        scanf("%d%d",&x,&y);
        add_edge(x,y);
        add_edge(y,x);
    }
    dfs1(r);
    dfs2(r,r);
    build(1,1,n);
    for(int i = 1; i <= m; i++){
        int op = 0,x = 0,y = 0,z = 0;
        scanf("%d%d",&op,&x);
        if(op == 1){
            scanf("%d%d",&y,&z);
            while(top[x] != top[y]){
                if(d[top[y]] > d[top[x]]) swap(x,y);
                update(1,id[top[x]],id[x],z);
                x = fa[top[x]];
            }
            if(d[x] > d[y]) swap(x,y);
                update(1,id[x],id[y],z);
        }
        if(op == 2){
            scanf("%d",&y);
            long long ans = 0;
            while(top[x] != top[y]){
                if(d[top[y]] > d[top[x]]) swap(x,y);
                ans = (ans + query(1,id[top[x]],id[x])) % p;
                x = fa[top[x]];
            }
            if(d[x] > d[y]) swap(x,y);
            ans = (ans + query(1,id[x],id[y]))% p;
            printf("%lld\n",ans);
        }
        if(op == 3){
            scanf("%d",&z);
            update(1,id[x],id[x] + siz[x] - 1,z);
        }
        if(op == 4){
            printf("%lld\n",query(1,id[x],id[x] + siz[x] - 1));
        } 
    }
    return 0;
}

[模板]树链剖分

标签:cpp   ++   +=   ret   i++   head   mes   int   its   

原文地址:https://www.cnblogs.com/FoxC/p/11256624.html

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