• BZOJ 1036 树链剖分


    题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1036

    题意:中文题目

    思路:树链剖分入门题, 剖分后用线段树维护即可。 

    #define _CRT_SECURE_NO_DEPRECATE
    #include<iostream>
    #include<cstring>
    #include<string>
    #include<algorithm>
    #include<stdio.h>
    #include<queue>
    #include<vector>
    #include<stack>
    #include<map>
    #include<set>
    #include<time.h>
    #include<cmath>
    #include<sstream>
    #include<assert.h>
    using namespace std;
    #define L(x) x<<1
    #define R(x) x<<1|1
    typedef long long int LL;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3fLL;
    const int MAXN = 30000 + 10;
    int val[MAXN], head[MAXN], tot, cnt;
    struct Edge{
        int to,next;
        Edge(int _to = 0, int _next = 0) :to(_to), next(_next){};
    }Edges[MAXN * 2];
    void add(int u, int v){
        Edges[tot].to = v;
        Edges[tot].next = head[u];
        head[u] = tot++;
    }
    int id[MAXN], son[MAXN], deep[MAXN], size[MAXN], fa[MAXN], reid[MAXN], top[MAXN];
    void Init(){
        tot = 0; cnt = 0;
        memset(head, -1, sizeof(head));
        memset(son, -1, sizeof(son));
    }
    void DFS1(int u, int p,int dep){
        fa[u] = p; size[u] = 1; deep[u] = dep;
        for (int i = head[u]; i != -1; i = Edges[i].next){
            if (Edges[i].to != p){
                DFS1(Edges[i].to, u,dep+1);
                size[u] += size[Edges[i].to];
                if (son[u] == -1 || size[Edges[i].to] > size[son[u]]){
                    son[u] = Edges[i].to;
                }
            }
        }
    }
    void DFS2(int u, int tp){
        id[u] = ++cnt; reid[id[u]] = u; top[u] = tp;
        if (son[u] == -1){ return; }
        DFS2(son[u], tp);
        for (int i = head[u]; i != -1; i = Edges[i].next){
            if (son[u] != Edges[i].to&&Edges[i].to != fa[u]){
                DFS2(Edges[i].to, Edges[i].to);
            }
        }
    }
    struct Node{
        int st, ed;
        LL sum, Max;
    }Seg[MAXN * 4];
    void Build(int l, int r, int k){
        Seg[k].st = l; Seg[k].ed = r; 
        if (l == r){
            Seg[k].sum = val[reid[l]]; Seg[k].Max = val[reid[l]];
            return;
        }
        int mid = (l + r) / 2;
        Build(l, mid, L(k)); Build(mid + 1, r, R(k));
        Seg[k].sum = Seg[L(k)].sum + Seg[R(k)].sum;
        Seg[k].Max = max(Seg[L(k)].Max, Seg[R(k)].Max);
    }
    void pushUp(int k){
        Seg[k].sum = Seg[L(k)].sum + Seg[R(k)].sum;
        Seg[k].Max = max(Seg[L(k)].Max, Seg[R(k)].Max);
    }
    void Modify(int pos, int k,int val){
        if (Seg[k].st ==Seg[k].ed){
            Seg[k].sum = val;
            Seg[k].Max = val;
            return;
        }
        if (pos <= Seg[L(k)].ed){
            Modify(pos, L(k), val);
        }
        else{
            Modify(pos, R(k), val);
        }
        pushUp(k);
    }
    LL QuerySum(int l, int r, int k){
        if (Seg[k].st == l&&Seg[k].ed == r){
            return Seg[k].sum;
        }
        LL sum = 0;
        if (r <= Seg[L(k)].ed){
            sum = QuerySum(l, r, L(k));
        }
        else if (l >= Seg[R(k)].st){
            sum = QuerySum(l, r, R(k));
        }
        else{
            sum = QuerySum(l, Seg[L(k)].ed, L(k)) + QuerySum(Seg[R(k)].st, r, R(k));
        }
        pushUp(k);
        return sum;
    }
    LL QuerySum(int u,int v){
        LL ans = 0;
        int f1 = top[u], f2 = top[v];
        while (f1!=f2){
            if (deep[f1] < deep[f2]){
                swap(f1, f2);
                swap(u, v);
            }
            ans += QuerySum(id[f1], id[u],1);
            u = fa[f1]; f1 = top[u];
        }
        if (deep[u] > deep[v]){
            swap(u, v);
        }
        ans += QuerySum(id[u],id[v], 1);
        return ans;
    }
    LL QueryMax(int l, int r, int k){
        if (Seg[k].st == l&&Seg[k].ed == r){
            return Seg[k].Max;
        }
        LL _Max = -inf;
        if (r <= Seg[L(k)].ed){
            _Max = QueryMax(l, r, L(k));
        }
        else if (l >= Seg[R(k)].st){
            _Max = QueryMax(l, r, R(k));
        }
        else{
            _Max = max(QueryMax(l, Seg[L(k)].ed, L(k)), QueryMax(Seg[R(k)].st, r, R(k)));
        }
        pushUp(k);
        return _Max;
    }
    LL QueryMax(int u,int v){
        LL ans = -inf;
        int f1 = top[u], f2 = top[v];
        while (f1 != f2){
            if (deep[f1] < deep[f2]){
                swap(f1, f2);
                swap(u, v);
            }
            ans = max(ans, QueryMax(id[f1], id[u], 1));
            u = fa[f1]; f1 = top[u];
        }
        if (deep[u] > deep[v]){
            swap(u, v);
        }
        ans = max(ans, QueryMax(id[u], id[v], 1));
        return ans;
    }
    int main(){
    //#ifdef kirito
    //    freopen("in.txt", "r", stdin);
    //    freopen("out.txt", "w", stdout);
    //#endif
    //    int start = clock();
        int n, m;
        while (~scanf("%d",&n)){
            Init();
            for (int i = 1; i < n; i++){
                int u, v;
                scanf("%d%d", &u, &v);
                add(u, v); add(v, u);
            }
            for (int i = 1; i <= n; i++){
                scanf("%d", &val[i]);
            }
            DFS1(1, 1, 0); DFS2(1, 1); Build(1, n, 1);
            scanf("%d", &m);
            while (m--){
                int x, a; char ope[10];
                scanf("%s%d%d", ope,&x,&a);
                switch (ope[1])
                {
                case 'M':printf("%lld
    ", QueryMax(x, a)); break;
                case 'S':printf("%lld
    ", QuerySum(x,a)); break;
                default: Modify(id[x], 1, a);  break;
                }
            }
        }
    //#ifdef LOCAL_TIME
    //    cout << "[Finished in " << clock() - start << " ms]" << endl;
    //#endif
        return 0;
    }
  • 相关阅读:
    Ubuntu Linux markdown编辑工具 typora 安装
    ref以及传值传址的理解
    3. 无重复字符的最长子串
    30. 串联所有单词的子串 (哈希+滑动窗口)
    525. 连续数组 (哈希表)
    438. 找到字符串中所有字母异位词 (滑动窗口)
    451、根据字符出现频率排序(哈希 加优先队列)
    743. 网络延迟时间
    310. 最小高度树
    8皇后问题
  • 原文地址:https://www.cnblogs.com/kirito520/p/6478493.html
Copyright © 2020-2023  润新知