Codeforces Round #381 (Div. 2) D Alyona and a tree DFS+前缀和+二分查找

D. Alyona and a tree

链接：

http://codeforces.com/contest/740/problem/D

题解：

对于在v的子树中的u，dis(u,v)其实就是dep[u]-dep[v]<=a[u]，移一下就是dep[u]-a[u]<=dep[v]，

那么直接dfs下去，二分找到第一个大于等于左边的dep[v]，那么相当于一个区间前缀和嘛，

然后dfs完统计答案并且把结点pop掉。

代码：

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
const ll mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double eps = 1e-7;
// head

const int maxn = 2e5 + 7;
vector<pair<int, ll> > e[maxn];
vector<pair<ll, int> > path;
ll a[maxn], dep[maxn], ans[maxn];

void dfs(int u){
    ans[u]++;
    int p = lower_bound(path.begin(), path.end(), make_pair(dep[u] - a[u], -1)) - path.begin();
    p--;
    if (p >= 0) ans[path[p].second]--;
     path.pb(mp(dep[u], u));
    for (int i = 0; i < e[u].size(); i++){
        int v = e[u][i].first;
        ll w = e[u][i].second;
        dep[v] = dep[u] + w;
        dfs(v);
        ans[u] += ans[v];
    }
    path.pop_back();
}

int main(){
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    rep(i, 1, n) cin >> a[i];
    rep(i, 2, n) {
        int p; ll w;
        cin >> p >> w;
        e[p].pb(mp(i, w));
    }
    dfs(1);
    rep(i, 1, n) cout << ans[i] - 1 << " ";
    return 0;
}

树状数组

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
const ll mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double eps = 1e-7;
// head

#define LL long long  
const int maxn = 2e5 + 5;
int a[maxn], in[maxn], out[maxn], c[maxn], ans[maxn];  //a-每个点的值，in-dfs该点的顺序，out-该点的子树最远的叶子  
LL dis[maxn]; //每个点到根的距离  
vector<int> g[maxn], w[maxn];  //g-建树，w-每条边的值  
int n, ord = 0;

struct node {
    int tp, id;
    LL v;
    node(){}
    node(int a, int b, LL c) {
        tp = a, id = b, v = c;
    }
} b[maxn * 2];

bool cmp(node a, node b) {
    if (a.v == b.v) return a.tp < b.tp;
    return a.v < b.v;
}

void dfs(int x, LL d) {  //dfs in-标记每个点遍历的顺序，out-该点的子树最远的叶子，dis-该点到根节点的距离  
    dis[x] = d;
    in[x] = ++ord;
    for (int i = 0; i < g[x].size(); i++) {
        dfs(g[x][i], d + w[x][i]);
    }
    out[x] = ord;
}

void add(int x) {
    while (x <= n) {
        c[x]++;
        x += x&-x; //lowbit直接写在函数里面，看起来简洁些  
    }
}

int query(int x) {
    int ans = 0;
    while (x) {
        ans += c[x];
        x -= x&-x;
    }
    return ans;
}

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 2; i <= n; i++) {
        int u, v;
        scanf("%d%d", &u, &v);
        g[u].pb(i);
        w[u].pb(v);
    }
    dfs(1, 0);
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
        b[++cnt] = node(1, i, dis[i] - a[i]);
        b[++cnt] = node(2, i, dis[i]);
    }    
    sort(b + 1, b + 1 + cnt, cmp);
    for (int i = 1; i <= cnt; i++) printf("%d  %d  %I64d
", b[i].tp, b[i].id, b[i].v);
    memset(c, 0, sizeof(c));
    for (int i = 1; i <= cnt; i++) {
        if (b[i].tp == 1) add(in[b[i].id]);
        else {
            ans[b[i].id] = query(out[b[i].id]) - query(in[b[i].id] - 1);
        }
    }
    for (int i = 1; i <= n; i++)
        printf("%d ", ans[i] - 1);  //得到的答案包含它本身，所以要减一  
    puts("");
}