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[HAOI2015]树上操作

时间:2017-12-24 20:07:59      阅读:162      评论:0      收藏:0      [点我收藏+]

标签:ring   fine   pac   find   out   void   namespace   targe   getch   

https://www.luogu.org/recordnew/show/5120068

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int N = 1e5 + 10;

#define gc getchar()
#define lson jd << 1
#define rson jd << 1 | 1
#define LL long long

LL ans;
int n, Ti, tim_clock, now = 1;
int head[N], tree[N], bef[N << 2], deep[N], data[N], fa[N], topp[N], siz[N], son[N], L[N], R[N];
struct Node{
    int u, v, nxt;
}G[N << 2];
struct Node_2{
    int l, r;
    LL w, f;
}T[N << 4];

inline int read(){
    int x = 0, f = 1; char c = gc;
    while(c < 0 || c > 9) {if(c == -) f = -1; c = gc;}
    while(c >= 0 && c <= 9) x = x * 10 + c - 0, c = gc;
    return x * f;
}

inline void swap(int & x, int & y){
    x ^= y; y ^= x; x ^= y;
}

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

void dfs_find_son(int u, int f, int dep){
    fa[u] = f; deep[u] = dep;
    siz[u] = 1;
    for(int i = head[u]; ~ i; i = G[i].nxt){
        int v = G[i].v;
        if(v != f) {
            dfs_find_son(v, u, dep + 1);
            siz[u] += siz[v];
            if(siz[v] > siz[son[u]]) son[u] = v;
        }
    }
}

void dfs_to_un(int u, int tp){
    topp[u] = tp;
    L[u] = ++ tim_clock;
    tree[u] = tim_clock;
    bef[tim_clock] = u;
    if(!son[u]){R[u] = tim_clock; return ;}
    dfs_to_un(son[u], tp);
    for(int i = head[u]; ~ i; i = G[i].nxt){
        int v = G[i].v;
        if(v != fa[u] && v != son[u]) dfs_to_un(v, v);
    }
    R[u] = tim_clock;
}

void down(int jd){
    int F = T[jd].f;
    T[lson].w += ((T[lson].r - T[lson].l + 1) * F); T[lson].f += F;
    T[rson].w += ((T[rson].r - T[rson].l + 1) * F); T[rson].f += F;
    T[jd].f = 0;
}

void build_tree(int l, int r, int jd){
    T[jd].l = l; T[jd].r = r;
    if(l == r){T[jd].w = data[bef[l]]; return ;}
    int mid = (l + r) >> 1;
    build_tree(l, mid, lson);
    build_tree(mid + 1, r, rson);
    T[jd].w = T[lson].w + T[rson].w;
}

void Poi_G(int l, int r, int jd, int x, int yj){
    if(l == r) {T[jd].w += yj; return ;}
    if(T[jd].f) down(jd);
    int mid = (l + r) >> 1;
    if(x <= mid) Poi_G(l, mid, lson, x, yj);
    else Poi_G(mid + 1, r, rson, x, yj);
    T[jd].w = T[lson].w + T[rson].w;
}

void Sec_G(int l, int r, int jd, int x, int y, int yj){
    if(x <= l && r <= y) {T[jd].w += ((r - l + 1) * yj); T[jd].f += yj; return ;}    if(T[jd].f) down(jd);
    int mid = (l + r) >> 1;
    if(x <= mid) Sec_G(l, mid, lson, x, y, yj);
    if(y > mid)  Sec_G(mid + 1, r, rson, x, y, yj);
    T[jd].w = T[lson].w + T[rson].w;
}

void Sec_A(int l, int r, int jd, int x, int y){
    if(x <= l && r <= y) {ans += T[jd].w; return ;}
    if(T[jd].f) down(jd);
    int mid = (l + r) >> 1;
    if(x <= mid) Sec_A(l, mid, lson, x, y);
    if(y > mid)  Sec_A(mid + 1, r, rson, x ,y);
}

inline LL Sec_A_I(int x, int y){
    int tp1 = topp[x], tp2 = topp[y]; LL answer = 0;
    while(tp1 != tp2){
        if(deep[tp1] < deep[tp2]) swap(x, y), swap(tp1, tp2);
        ans = 0;
        Sec_A(1, n, 1, tree[tp1], tree[x]);
        answer += ans;
        x = fa[tp1];
        tp1 = topp[x];
    }
    ans = 0;
    if(deep[x] < deep[y]) swap(x, y);
    Sec_A(1, n, 1, tree[y], tree[x]);
    answer += ans;
    return answer;
}

void debug(){
    for(int i = 1; i <= n; i ++) cout << tree[i] << " ";
    cout << endl;
    exit(0);
}

int main()
{
    n = read(); Ti = read();
    for(int i = 1; i <= n; i ++) data[i] = read();
    for(int i = 1; i <= n; i ++) head[i] = -1;
    for(int i = 1; i < n ; i ++){
        int u = read(), v = read();
        add(u, v); add(v, u);
    }
    
    dfs_find_son(1, 0, 1);
    dfs_to_un(1, 1);
    build_tree(1, n, 1);
    while(Ti --){
        int how = read();
        if(how == 1){int x = read(), a = read(); Poi_G(1, n, 1, tree[x], a);}
        else if(how == 2) {int x = read(), a = read(); Sec_G(1, n, 1, L[x], R[x], a);}
        else {int x = read(); printf("%lld\n", Sec_A_I(x, 1));}
    }
    
    return 0;
}
/*
5 5
1 2 3 4 5
1 2
1 4
2 3
2 5
3 3
1 2 1
3 5
2 1 2
3 3
*/

 

[HAOI2015]树上操作

标签:ring   fine   pac   find   out   void   namespace   targe   getch   

原文地址:http://www.cnblogs.com/shandongs1/p/8098749.html

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