[hihicoder][Offer收割]编程练习赛47

删除树节点

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<functional>
#include<math.h>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef queue<int> QI;


void makedata() {
    freopen("input.txt", "w", stdout);
    fclose(stdout);
}

int w[110000], fa[110000], n, k;
VI G[110000];
void dfs(int x, int f) {
    fa[x] = f;

    if(w[x] < k) {
        fa[x] = -1;

        for(int i = 0; i < G[x].size(); i++) {
            int y = G[x][i];
            dfs(y, f);
        }
    } else {
        for(int i = 0; i < G[x].size(); i++) {
            int y = G[x][i];
            dfs(y, x);
        }
    }
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(0), cin.tie(0);
    int root;
    cin >> n >> k;

    for(int i = 1; i <= n; i++) cin >> w[i];

    for(int i = 1; i <= n; i++) {
        cin >> fa[i];
        G[fa[i]].push_back(i);

        if(fa[i] == 0) root = i;
    }

    dfs(root, 0);

    for(int i = 1; i <= n; i++) cout << fa[i] << ' ';

    return 0;
}

鱼雷射击

mei yi si lan de xie le

 公平分队II

一个人拿到最大的两个，另一个拿到最小的两个，剩下的平均分配。

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<functional>
#include<math.h>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef queue<int> QI;


void makedata() {
    freopen("input.txt", "w", stdout);
    fclose(stdout);
}


int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(0), cin.tie(0);
    int n;
    cin >> n;
    lint ans = 0;
    for (int i = 3; i < 2 * n - 1; i++) ans += i;
    ans /= 2;
    ans += 4 * n - 1;
    if (n == 1) ans = 2;
    cout << ans << endl;
    return 0;
}

数组区间

依次计算每个数字在多少个区间中出现，乘起来求和。

p[i][j]表示第i个数字左边第j个比它大的数字的位置，q[i][j]表示第i个数字右边第j个比它大的数字的位置。因为数据中每个数字都不相同，所以按从小到大的顺序计算每个数字的p和q，只需直接取其左右的连续数字即可，计算完成后将该数字删除。

对于第i个数字出现区间数的计算，枚举其左边比它大的数字j，左边界的可能种类数为p[i][j] - p[i][j + 1]，右边界的可能种类数为q[i][k - j] - q[i][0]。

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<functional>
#include<math.h>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef queue<int> QI;


void makedata() {
    freopen("input.txt", "w", stdout);
    fclose(stdout);
}

lint a[110000], pre[110000], nex[110000];
lint p[110000][60], q[110000][60];
vector<PII> v;

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(0), cin.tie(0);
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; i++) cin >> a[i];
    v.clear();
    for (int i = 1; i <= n; i++) v.push_back(make_pair(a[i], i));
    sort(v.begin(), v.end());
    for (int i = 1; i <= n; i++) pre[i] = i - 1, nex[i] = i + 1;
    for (int i = 0; i < v.size(); i++) {
        int id = v[i].second, ptr;
        p[id][0] = q[id][0] = id;
        ptr = id;
        for (int i = 1; i <= k; i++) {
            if (pre[ptr] > 0) {
                p[id][i] = pre[ptr];
                ptr = pre[ptr];
            } else break;
        }
        ptr = id;
        for (int i = 1; i <= k; i++) {
            if (nex[ptr] <= n) {
                q[id][i] = nex[ptr];
                ptr = nex[ptr];
            } else break;
        }
        nex[pre[id]] = nex[id];
        pre[nex[id]] = pre[id];
    }
    lint ans = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= k; j++) {
            if (q[i][j] == 0) q[i][j] = n + 1;
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < k; j++) {
            if (p[i][j] == 0) break;
            ans += (p[i][j] - p[i][j + 1]) * (q[i][k - j] - q[i][0]) * a[i];
        }
    }
    cout << ans << endl;
    return 0;
}