Codeforces Round #200 (Div. 1) D. Water Tree(dfs序加线段树)

思路:

dfs序其实是很水的东西。  和树链剖分一样, 都是对树链的hash。

该题做法是:每次对子树全部赋值为1,对一个点赋值为0,查询子树最小值。

该题需要注意的是:当我们对一棵子树全都赋值为1的时候, 我们要查询一下赋值前子树最小值是不是0, 如果是的话, 要让该子树父节点变成0, 否则变0的信息会丢失。

细节参见代码:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <ctime>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const double eps = 1e-6;
const double PI = acos(-1);
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
// & 0x7FFFFFFF
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 500000+10;
int T,n,m,tot=0,in[maxn],out[maxn],minv[maxn<<2],setv[maxn<<2],pre[maxn];
vector<int> g[maxn];
void dfs(int u, int fa) {
    in[u] = ++tot;
    pre[u] = fa;
    int len = g[u].size();
    for(int i = 0; i < len; i++) {
        int v = g[u][i];
        if(v == fa) continue;
        dfs(v, u);
    }
    out[u] = tot;
}
void pushup(int o) {
    minv[o] = min(minv[o<<1], minv[o<<1|1]);
}
void pushdown(int l, int r, int o) {
    if(setv[o] != -1) {
        setv[o<<1] = setv[o<<1|1] = setv[o];
        minv[o<<1] = minv[o<<1|1] = setv[o];
        setv[o] = -1;
    }
}
void build(int l, int r, int o) {
    minv[o] = 0;
    setv[o] = -1;
    if(l == r) return ;
    int mid = (l + r) >> 1;
    build(l, mid, o<<1);
    build(mid+1, r, o<<1|1);
}
void update(int L, int R, int v, int l, int r, int o) {
    if(L <= l && r <= R) {
        minv[o] = v;
        setv[o] = v;
        return ;
    }
    int mid = (l + r) >> 1;
    pushdown(l, r, o);
    if(L <= mid) update(L, R, v, l, mid, o<<1);
    if(mid < R) update(L, R, v, mid+1, r, o<<1|1);
    pushup(o);
}
int query(int L, int R, int l, int r, int o) {
    if(L <= l && r <= R) {
        return minv[o];
    }
    int mid = (l + r) >> 1;
    pushdown(l, r, o);
    int ans = INF;
    if(L <= mid) ans = min(ans, query(L, R, l, mid, o<<1));
    if(mid < R) ans = min(ans, query(L, R, mid+1, r, o<<1|1));
    pushup(o);
    return ans;
}
int main() {
    scanf("%d", &n);
    for(int i = 1; i < n; i++) {
        int u, v; scanf("%d%d", &u, &v);
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1, 0);
    build(1, n, 1);
    int q; scanf("%d", &q);
    while(q--) {
        int id, v; scanf("%d%d", &id, &v);
        if(id == 1) {
            int tmp = query(in[v], out[v], 1, n, 1);
            update(in[v], out[v], 1, 1, n, 1);
            if(!tmp && pre[v]) update(in[pre[v]], in[pre[v]], 0, 1, n, 1);
        }
        else if(id == 2) {
            update(in[v], in[v], 0, 1, n, 1);
        }
        else {
            printf("%d\n", query(in[v], out[v], 1, n, 1));
        }
    }
    return 0;
}

posted on 2016-12-07 19:08  如果蜗牛有爱情  阅读(472)  评论(0编辑  收藏  举报

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