bzoj4034 线段树+dfs序

https://www.lydsy.com/JudgeOnline/problem.php?id=4034

 

有一棵点数为 N 的树,以点 1 为根,且树点有边权。然后有 M 个
操作,分为三种:
操作 1 :把某个节点 x 的点权增加 a 。
操作 2 :把某个节点 x 为根的子树中所有点的点权都增加 a 。
操作 3 :询问某个节点 x 到根的路径中所有点的点权和。
 
dfs序真是一个奥妙重重的东西
首先求出dfs序,操作1显然是一个单点修改,操作2显然是一个区间修改,我们看到我们要询问的事实上是一条链的和,我们就不能简单地进行单点修改和区间修改,事实上1到x的点在dfs序上可以视为一个栈,第一次出现是进栈,第二次出现时出栈,我们只要将进栈的点权为正,出栈的点权为负,线段树进行一波区间修改和区间查询即可。
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
inline int read(){int now=0;register char c=getchar();for(;!isdigit(c);c=getchar());
for(;isdigit(c);now=now*10+c-'0',c=getchar());return now;}
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
const double eps = 1e-9;
const int maxn = 4e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
LL val[maxn];
struct Edge{
    int to,next;
}edge[maxn * 2];
int head[maxn],tot;
void init(){Mem(head,-1); tot = 0;}
void add(int u,int v){edge[tot].to = v;edge[tot].next = head[u];head[u] = tot++;}
PII Index[maxn * 2];
PII pos[maxn];
int id = 0;
void dfs(int t,int fa){
    Index[++id].fi = t;
    Index[id].se = 1;
    pos[t].fi = id;
    for(int i = head[t]; ~i; i = edge[i].next){
        int v = edge[i].to;
        if(v == fa) continue;
        dfs(v,t);
    }
    Index[++id].fi = t;
    Index[id].se = -1;
    pos[t].se = id;
}
struct Tree{
    int l,r;
    int up,down;
    LL sum,lazy;
}tree[maxn << 2];
void Pushup(int t){
    tree[t].sum = tree[t << 1].sum + tree[t << 1 | 1].sum;
}
void Build(int t,int l,int r){
    tree[t].l = l; tree[t].r = r;
    tree[t].up = tree[t].down = tree[t].lazy = 0;
    if(l == r){
        if(Index[l].se == 1){
            tree[t].sum = val[Index[l].fi];
            tree[t].up++;
        }else{
            tree[t].sum = -val[Index[l].fi];
            tree[t].down++;
        }
        return ;
    }
    int m = (l + r) >> 1;
    Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r);
    Pushup(t);
    tree[t].up = tree[t << 1].up + tree[t << 1 | 1].up;
    tree[t].down = tree[t << 1].down + tree[t << 1 | 1].down; 
}
void Pushdown(int t){
    if(tree[t].lazy){
        tree[t << 1].sum += (tree[t << 1].up - tree[t << 1].down) * tree[t].lazy;
        tree[t << 1 | 1].sum += (tree[t << 1 | 1].up - tree[t << 1 | 1].down) * tree[t].lazy;
        tree[t << 1].lazy += tree[t].lazy; tree[t << 1 | 1].lazy += tree[t].lazy;
        tree[t].lazy = 0;
    }
}
void update(int t,int l,int r,LL a){
    if(l <= tree[t].l && tree[t].r <= r){
        tree[t].sum += (tree[t].up - tree[t].down) * a;
        tree[t].lazy += a;
        return;
    }
    Pushdown(t);
    int m = (tree[t].l + tree[t].r) >> 1;
    if(r <= m) update(t << 1,l,r,a);
    else if(l > m) update(t << 1 | 1,l,r,a);
    else{
        update(t << 1,l,m,a);
        update(t << 1 | 1,m + 1,r,a);
    }
    Pushup(t);
}
LL query(int t,int l,int r){
    if(l > r) return 0;
    if(l <= tree[t].l && tree[t].r <= r){
        return tree[t].sum;
    }
    Pushdown(t);
    int m = (tree[t].l + tree[t].r) >> 1;
    if(r <= m) return query(t << 1,l,r);
    else if(l > m) return query(t << 1 | 1,l,r);
    else{
        return query(t << 1,l,m) + query(t << 1 | 1,m + 1,r);
    }
}
int main()
{
    Sca2(N,M); init();
    For(i,1,N) Scl(val[i]);
    For(i,1,N - 1){ int u,v; Sca2(u,v); add(u,v); add(v,u);}
    int root = 1; id = 0;
    dfs(root,-1);
    Build(1,1,2 * N); 
    while(M--){
        int op = read();
        if(op == 1){
            int x; Sca(x); LL a; Scl(a);
            update(1,pos[x].fi,pos[x].fi,a); update(1,pos[x].se,pos[x].se,a);
        }else if(op == 2){
            int x; Sca(x); LL a; Scl(a);
            update(1,pos[x].fi,pos[x].se,a);
        }else{
            int x; Sca(x);
            Prl(query(1,1,pos[x].fi));
        }
    }
    #ifdef VSCode
    system("pause");
    #endif
    return 0;
}

 

posted @ 2018-10-24 13:11  Hugh_Locke  阅读(223)  评论(0编辑  收藏  举报