模板——树链剖分

放个板子而已。。

#include<stdio.h>
#include<algorithm>
#define ls root<<1
#define rs root<<1|1
using namespace std;
typedef long long ll;

int n,m,r;ll res,p;

const int MAXN = 100005;
ll a[MAXN];
struct edge{
    int v,nxt;
}e[MAXN<<1];

struct TREE{
    int lazy,l,r,val;
}tree[MAXN*4];
int head[MAXN];int cnt=0;int dep[MAXN];int fa[MAXN];int son[MAXN];int size[MAXN];int top[MAXN];int newval[MAXN];int id[MAXN];
int tot=0;

inline void add(int u,int v){
    e[++cnt].v = v;e[cnt].nxt = head[u];head[u] = cnt;
}

inline void pushdown(int root,int len){
    tree[ls].lazy += tree[root].lazy;
    tree[rs].lazy += tree[root].lazy;
    tree[ls].val += tree[root].lazy * (len-(len>>1));
    tree[rs].val += tree[root].lazy * (len>>1);
    tree[ls].val %= p;
    tree[rs].val %= p;
    tree[root].lazy = 0;
}

inline void update(int root,int L,int R,int l,int r,ll k){
    if(l <= L && R <= r){
        tree[root].lazy += k;
        tree[root].val += (R-L+1)*k;
    }
    else{
        int mid = (L + R) >> 1;
        if(tree[root].lazy) pushdown(root,R-L+1);
        if(l <= mid) update(ls,L,mid,l,r,k);
        if(r >= mid+1) update(rs,mid+1,R,l,r,k);
        tree[root].val = tree[ls].val + tree[rs].val;
        tree[root].val %= p;
    }
}

inline void query(int root,int L,int R,int l,int r){
    if(l <= L && R <= r){
        res += tree[root].val;
        res %= p;
        return ;
    }
    else{
        int mid = (L + R) >> 1;
        if(tree[root].lazy) pushdown(root,R-L+1);
        if(l <= mid) query(ls,L,mid,l,r);
        if(r >= mid+1) query(rs,mid+1,R,l,r);
    }
}

//////////////////////////线段树分割线////////////////////////////////////

inline void dfs1(int root,int father,int depth){
    dep[root] = depth;
    fa[root] = father;
    size[root] = 1;
    for(int i=head[root];i;i=e[i].nxt){
        int v = e[i].v;
        if(v == father) continue;
        dfs1(v,root,depth+1);
        size[root] += size[v];
        if(!son[root] || size[v] > size[son[root]]){
            son[root] = v;
        }
    }
}

inline void dfs2(int u,int nowtop){
    id[u] = ++tot;
    top[u] = nowtop;
    newval[tot] = a[u];
    if(!son[u]) return ;
    dfs2(son[u] , nowtop);
    for(int i=head[u];i;i=e[i].nxt){
        int v = e[i].v;
        if(v == fa[u] || v == son[u]) continue;
        dfs2(v,v);
    }
}

inline void build(int root,int l,int r){
    if(l == r){
        tree[root].val = newval[l];
        tree[root].val %= p;
        return ;
    }
    int mid = (l + r) >> 1;
    build(ls,l,mid);
    build(rs,mid+1,r);
    tree[root].val = tree[ls].val + tree[rs].val;
    tree[root].val %= p;
}

inline void addrange(int l,int r,ll num){
    while(top[l] ^ top[r]){
        if(dep[top[l]] < dep[top[r]]) swap(l,r);
        update(1,1,n,id[top[l]],id[l],num);
        l = fa[top[l]];
    }
    if(dep[l] > dep[r]) swap(l,r);
    update(1,1,n,id[l],id[r],num);
}

inline int queryrange(int l,int r){
    int ans = 0;
    while(top[l] ^ top[r]){
        if(dep[top[l]] < dep[top[r]]) swap(l,r);
        res=0;
        query(1,1,n,id[top[l]],id[l]);
        ans += res;
        ans %= p;
        l = fa[top[l]];
    }
    if(dep[l] > dep[r]) swap(l,r);
    res=0;
    query(1,1,n,id[l],id[r]);
    ans += res;
    return ans%p;
}

int main(){
    scanf("%d%d%d%lld",&n,&m,&r,&p);
    for(int i=1;i<=n;++i) scanf("%lld",&a[i]);
    for(int i=1;i<n;++i){
        int u,v;scanf("%d%d",&u,&v);
        add(u,v);add(v,u);
    }
    dfs1(r,0,1);
    dfs2(r,r);
    build(1,1,n);
    for(;m;--m){
        int k;scanf("%d",&k);
        switch (k){
            case 1:{
                int x,y;ll z;
                scanf("%d%d%lld",&x,&y,&z);
                z %= p;
                addrange(x,y,z);
                break;
            }
            case 2:{
                int x,y;
                scanf("%d%d",&x,&y);
                printf("%d\n",queryrange(x,y));
                break;
            }
            case 3:{
                int x;ll z;
                scanf("%d%lld",&x,&z);
                update(1,1,n,id[x],id[x]+size[x]-1,z);
                break;
            }
            case 4:{
                int x;scanf("%d",&x);
                res = 0;
                query(1,1,n,id[x],id[x]+size[x]-1);
                printf("%lld\n",res%p);
                break;
            }
        }
    }
    return 0;
}
posted @ 2018-08-15 00:31  lajioj  阅读(118)  评论(0编辑  收藏  举报