zoukankan      html  css  js  c++  java
  • [Offer收割]编程练习赛108

    https://hihocoder.com/contest/offers108/problems

    A 再买优惠

    暴力或二分一下

    #include <bits/stdc++.h>
    #define DBG(x) cerr << #x << " = " << x << endl
    
    using namespace std;
    typedef long long LL;
    
    struct Price {
        int a, b;
        bool operator < (const Price &rhs) const {
            return this->a < rhs.a;
        }
    };
    
    int main(int argc, char **argv) {
        int n, c;
        while (~scanf("%d%d", &n, &c)) {
            vector<Price> vec;
            for (int i = 0; i < n; ++i) {
                Price p;
                scanf("%d%d", &p.a, &p.b);
                vec.push_back(p);
            }
            sort(vec.begin(), vec.end());
            int pos = upper_bound(vec.begin(), vec.end(), Price{c}) - vec.begin();
            if (pos == n) {
                printf("%d
    ", vec[pos - 1].b);
            } else {
                printf("%d %d %d
    ", vec[pos - 1].b, vec[pos].a - c, vec[pos].b);
            }
        }
        return 0;
    }
    
    
    /**
    3 40
    20 10
    50 25
    100 50
    
    10 10 25
    */
    

    B 树与落叶

    DFS + 维护 + 二分

    #include <bits/stdc++.h>
    
    #define DBG(x) cerr << #x << " = " << x << endl
    
    using namespace std;
    typedef long long LL;
    
    const int N = 1000000 + 16;
    
    vector<int> G[N];
    int max_depth, drop_time[N], total[N], drop_cnt[N];
    
    void init(int n) {
        for (int i = 0; i <= n; ++i) {
            G[i].clear();
            total[i] = 0;
            drop_cnt[i] = 0;
        }
        max_depth = 0;
    }
    
    void dfs(int u, int fa, int depth) {
        drop_time[u] = 1;
        max_depth = max(max_depth, depth);
        for (auto v : G[u]) {
            if (v == fa) continue;
            dfs(v, u, depth + 1);
            drop_time[u] = max(drop_time[u], drop_time[v] + 1);
        }
    }
    
    int main(int argc, char **argv) {
        int n, m;
        while (~scanf("%d%d", &n, &m)) {
            init(n);
            int root = 1;
            for (int i = 1; i <= n; ++i) {
                int f;
                scanf("%d", &f);
                if (f == 0) {
                    root = i;
                    continue;
                }
                G[f].push_back(i);
            }
            dfs(root, -1, 1);
    
            int len = drop_time[root] + 1;
            for (int i = 1; i <= n; i++)
                drop_cnt[drop_time[i]]++;
            total[0] = n;
            for (int i = 1; i < len; i++) {
                total[i] = total[i - 1] - drop_cnt[i];
            }
    
            while (m--) {
                int d;
                scanf("%d", &d);
                int pos = lower_bound(total, total + len, d, greater<int>()) - total;
                if (pos == len) {
                    printf("%d
    ", len);
                } else if (pos == 0) {
                    printf("%d
    ", 1);
                } else {
                    if (abs(total[pos - 1] - d) <= abs(total[pos] - d)) {
                        printf("%d
    ", pos);
                    } else {
                        printf("%d
    ", pos + 1);
                    }
                }
            }
        }
        return 0;
    }
    
    /**
    5 2
    2 0 2 3 3
    4 0
    
    1
    4
    */
    

    C 树上的最短边

    我是用树链剖分做的

    #include <bits/stdc++.h>
    #define DBG(x) cerr << #x << " = " << x << endl
    #define lrrt int L, int R, int rt
    #define lson L, mid, rt << 1
    #define rson mid + 1, R, rt << 1 | 1
    #define iall 1, n, 1
    
    using namespace std;
    typedef long long LL;
    
    const int N = 100000 + 16;
    
    int tree[N << 2];
    vector<int> G[N];
    int n, m, ans[N];
    int D[N], F[N], sz[N], son[N];
    int W[N], _W[N], top[N], dfs_cnt;
    void dfs(int u, int fa, int dep) {
        D[u] = dep;
        F[u] = fa;
        son[u] = 0;
        sz[u] = 1;
        for(int i = 0; i < G[u].size(); ++i) {
            int v = G[u][i];
            if(v != fa) {
                dfs(v, u, dep + 1);
                sz[u] += sz[v];
                if(son[u] == 0 || sz[v] > sz[son[u]]) son[u] = v;
            }
        }
    }
    void dfs_hash(int u, int tp, int fa) {
        W[u] = ++dfs_cnt;
        _W[W[u]] = u;
        top[u] = tp;
        if(son[u]) dfs_hash(son[u], tp, u);
        for(int i = 0; i < G[u].size(); ++i) {
            int v = G[u][i];
            if(v == son[u] || v == fa) continue;
            dfs_hash(v, v, u);
        }
    }
    void build(lrrt) {
        if(L == R) {
            tree[rt] = 0x3F3F3F3F;
        } else {
            int mid = L + R >> 1;
            build(lson);
            build(rson);
            tree[rt] = min(tree[rt << 1], tree[rt << 1 | 1]);
        }
    }
    void add(lrrt, int pos, int v) {
        if(L == R) {
            tree[rt] = v;
        } else {
            int mid = L + R >> 1;
            if(pos <= mid) add(lson, pos, v);
            else add(rson, pos, v);
            tree[rt] = min(tree[rt << 1], tree[rt << 1 | 1]);
        }
    }
    int query(lrrt, int x, int y) {
        if(x <= L && R <= y) return tree[rt];
        else {
            int mid = L + R >> 1;
            if(x > mid) return query(rson, x, y);
            else if(y <= mid) return query(lson, x, y);
            else return min(query(lson, x, y), query(rson, x, y));
        }
    }
    int query(int u, int v) {
        int res = 0x3F3F3F3F;
        while(top[u] != top[v]) {
            if(D[top[u]] < D[top[v]]) swap(u, v);
            res = min(res, query(iall, W[top[u]], W[u]));
            u = F[top[u]];
        }
        if(D[u] > D[v]) swap(u, v);
        if (u != v) {
            res = min(res, query(iall, W[son[u]], W[v]));
        }
        return res;
    }
    void add(int u, int val) {
        add(iall, W[u], val);
    }
    void init() {
        for(int i = 1; i <= n; ++i) G[i].clear();
        dfs_cnt = 0;
    }
    
    int pa[N], wc[N];
    
    int main(int argc, char **argv) {
        while(~scanf("%d%d", &n, &m)) {
            init();
            int root = 1;
            for (int i = 1; i <= n; ++i) {
                scanf("%d%d", &pa[i], &wc[i]);
                if (pa[i] == 0) {
                    root = i;
                    continue;
                }
                G[pa[i]].push_back(i);
                G[i].push_back(pa[i]);
            }
            dfs(root, -1, 0); dfs_hash(root, root, -1);
            build(iall);
            for (int i = 1; i <= n; ++i) {
                if (wc[i] == 0)
                    continue;
                add(i, wc[i]);
            }
            while (m--) {
                int u, v;
                scanf("%d%d", &u, &v);
                printf("%d
    ", query(u, v));
            }
        }
        return 0;
    }
    
    /**
    5 2
    2 1
    0 0
    2 2
    3 3
    3 4
    4 1
    2 5
    
    1
    2
    */
    

    D 01匹配

    线段树

    #include <bits/stdc++.h>
    #define DBG(x) cerr << #x << " = " << x << endl
    
    #define lrrt int L, int R, int rt
    #define lson L, mid, rt << 1
    #define rson mid + 1, R, rt << 1 | 1
    #define iall 1, n, 1
    
    using namespace std;
    typedef long long LL;
    
    const int N = 300000 + 16;
    
    struct Node {
        int ans, free_one, free_zero;
        Node() {}
        Node(int ans, int free_one, int free_zero):
            ans(ans), free_one(free_one), free_zero(free_zero) {}
        Node operator &(const Node &rhs) const {
            Node ret;
            ret.ans = this->ans + rhs.ans;
            int add = min(this->free_one, rhs.free_zero);
            ret.ans += add;
            ret.free_one = this->free_one + rhs.free_one - add;
            ret.free_zero = this->free_zero + rhs.free_zero - add;
            return ret;
        }
    };
    
    int a[N];
    Node tree[N << 2];
    
    void build(lrrt) {
        tree[rt].ans = 0;
        tree[rt].free_zero = 0;
        tree[rt].free_one = 0;
        if (L == R) {
            if (a[L]) tree[rt].free_one = 1;
            else tree[rt].free_zero = 1;
            return;
        }
        int mid = (L + R) / 2;
        build(lson);
        build(rson);
        tree[rt] = tree[rt << 1] & tree[rt << 1 | 1];
    }
    
    Node query(lrrt, int x, int y) {
        if (x <= L && R <= y) {
            return tree[rt];
        }
        int mid = (L + R) / 2;
        if (y <= mid) return query(lson, x, y);
        else if (x > mid) return query(rson, x, y);
        return query(lson, x, y) & query(rson, x, y);
    }
    
    int main(int argc, char **argv) {
        int n;
        while (~scanf("%d", &n)) {
            for (int i = 1; i <= n; i++)
                scanf("%d", &a[i]);
            build(iall);
            int m;
            scanf("%d", &m);
            while (m--) {
                int l, r;
                scanf("%d%d", &l, &r);
                printf("%d
    ", query(iall, l, r).ans);
            }
        }
        return 0;
    }
    
    /**
    10
    0 1 0 1 0 0 1 0 1 0
    100
    1 10
    
    4
    */
    
  • 相关阅读:
    高盛、沃尔玛 题做出来还挂了的吐槽
    amazon师兄debrief
    到所有人家距离之和最短的中点 296. Best Meeting Point
    问问题没人回答的情况怎么办终于有解了
    找名人 277. Find the Celebrity
    数组生存游戏 289. Game of Life
    547. Number of Provinces 省份数量
    428. Serialize and Deserialize Nary Tree 序列化、反序列化n叉树
    alias别名简介和使用
    面试官:线程池执行过程中遇到异常会发生什么,怎样处理? Vincent
  • 原文地址:https://www.cnblogs.com/ToRapture/p/11767308.html
Copyright © 2011-2022 走看看