• [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;
    }
    View Code

    鱼雷射击

    mei yi si lan de xie le
    View Code

     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;
    }
    View Code

    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;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/dramstadt/p/8459931.html
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