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

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

        for(int i = ; i < G[x].size(); i++) {
            int y = G[x][i];
            dfs(y, f);
        }
    } else {
        for(int i = ; 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(), cin.tie();
    int root;
    cin >> n >> k;

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

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

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

    dfs(root, );

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

    return ;
}

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(), cin.tie();
    int n;
    cin >> n;
    lint ans = ;
    for (int i = ; i <  * n - ; i++) ans += i;
    ans /= ;
    ans +=  * n - ;
    if (n == ) ans = ;
    cout << ans << endl;
    return ;
}

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

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