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  • BZOJ 1036 [ZJOI2008]树的统计Count (树链剖分)(线段树单点修改)

    [ZJOI2008]树的统计Count

    Time Limit: 10 Sec  Memory Limit: 162 MB
    Submit: 14968  Solved: 6079
    [Submit][Status][Discuss]

    Description

      一棵树上有n个节点,编号分别为1到n,每个节点都有一个权值w。我们将以下面的形式来要求你对这棵树完成
    一些操作: I. CHANGE u t : 把结点u的权值改为t II. QMAX u v: 询问从点u到点v的路径上的节点的最大权值 I
    II. QSUM u v: 询问从点u到点v的路径上的节点的权值和 注意:从点u到点v的路径上的节点包括u和v本身

    Input

      输入的第一行为一个整数n,表示节点的个数。接下来n – 1行,每行2个整数a和b,表示节点a和节点b之间有
    一条边相连。接下来n行,每行一个整数,第i行的整数wi表示节点i的权值。接下来1行,为一个整数q,表示操作
    的总数。接下来q行,每行一个操作,以“CHANGE u t”或者“QMAX u v”或者“QSUM u v”的形式给出。
    对于100%的数据,保证1<=n<=30000,0<=q<=200000;中途操作中保证每个节点的权值w在-30000到30000之间。

    Output

      对于每个“QMAX”或者“QSUM”的操作,每行输出一个整数表示要求输出的结果。

    Sample Input

    4
    1 2
    2 3
    4 1
    4 2 1 3
    12
    QMAX 3 4
    QMAX 3 3
    QMAX 3 2
    QMAX 2 3
    QSUM 3 4
    QSUM 2 1
    CHANGE 1 5
    QMAX 3 4
    CHANGE 3 6
    QMAX 3 4
    QMAX 2 4
    QSUM 3 4

    Sample Output

    4
    1
    2
    2
    10
    6
    5
    6
    5
    16
    【分析】很简单的一道树链剖分。唯一要注意的就是负数问题。。。
     
    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <time.h>
    #include <string>
    #include <map>
    #include <stack>
    #include <vector>
    #include <set>
    #include <queue>
    #define met(a,b) memset(a,b,sizeof a)
    #define pb push_back
    #define lson(x) ((x<<1))
    #define rson(x) ((x<<1)+1)
    using namespace std;
    typedef long long ll;
    const int N=2e5+50;
    const int M=N*N+10;
    int dep[N],siz[N],fa[N],id[N],son[N],val[N],top[N],c[N]; //top 最近的重链父节点
    int num,n,m;
    vector<int> v[N];
    struct tree {
        int x,y;
        void read() {
            scanf("%d%d",&x,&y);
        }
    };
    tree e[N];
    void dfs1(int u, int f, int d) {
        dep[u] = d;
        siz[u] = 1;
        son[u] = 0;
        fa[u] = f;
        for (int i = 0; i < v[u].size(); i++) {
            int ff = v[u][i];
            if (ff == f) continue;
            dfs1(ff, u, d + 1);
            siz[u] += siz[ff];
            if (siz[son[u]] < siz[ff])
                son[u] = ff;
        }
    }
    void dfs2(int u, int tp) {
        top[u] = tp;
        id[u] = ++num;
        if (son[u]) dfs2(son[u], tp);
        for (int i = 0; i < v[u].size(); i++) {
            int ff = v[u][i];
            if (ff == fa[u] || ff == son[u]) continue;
            dfs2(ff, ff);
        }
    }
    
    struct Tree {
        int l,r,val,sum;
    };
    Tree tree[4*N];
    void pushup(int x) {
        tree[x].val = max(tree[lson(x)].val, tree[rson(x)].val);
        tree[x].sum=tree[lson(x)].sum+tree[rson(x)].sum;
    }
    
    void build(int l,int r,int v) {
        tree[v].l=l;
        tree[v].r=r;
        if(l==r) {
            tree[v].val = val[l];
            tree[v].sum=val[l];
            return ;
        }
        int mid=(l+r)>>1;
        build(l,mid,v*2);
        build(mid+1,r,v*2+1);
        pushup(v);
    }
    void update(int o,int v,int val) { //log(n)
        if(tree[o].l==tree[o].r) {
            tree[o].val =tree[o].sum= val;
            return ;
        }
        int mid = (tree[o].l+tree[o].r)/2;
        if(v<=mid)
            update(o*2,v,val);
        else
            update(o*2+1,v,val);
        pushup(o);
    }
    int queryMax(int x,int l, int r) {
        if (tree[x].l >= l && tree[x].r <= r) {
            return tree[x].val;
        }
        int mid = (tree[x].l + tree[x].r) / 2;
        int ans = -100000000;
        if (l <= mid) ans = max(ans, queryMax(lson(x),l,r));
        if (r > mid) ans = max(ans, queryMax(rson(x),l,r));
        return ans;
    }
    
    int Qmax(int u, int v) {
        int tp1 = top[u], tp2 = top[v];
        int ans = -100000000;
        while (tp1 != tp2) {
            if (dep[tp1] < dep[tp2]) {
                swap(tp1, tp2);
                swap(u, v);
            }
            ans = max(queryMax(1,id[tp1], id[u]), ans);
            u = fa[tp1];
            tp1 = top[u];
        }
        //if (u == v) return ans;
        if (dep[u] > dep[v]) swap(u, v);
        ans = max(queryMax(1,id[u], id[v]), ans);
        return ans;
    }
    int querySum(int x,int l,int r){
        if (tree[x].l >= l && tree[x].r <= r) {
            return tree[x].sum;
        }
        int mid = (tree[x].l + tree[x].r) / 2;
        int ans = 0;
        if (l <= mid) ans += querySum(lson(x),l,r);
        if (r > mid) ans += querySum(rson(x),l,r);
        return ans;
    }
    int Qsum(int u,int v){
        int tp1 = top[u], tp2 = top[v];
        int ans = 0;
        while (tp1 != tp2) {
            if (dep[tp1] < dep[tp2]) {
                swap(tp1, tp2);
                swap(u, v);
            }
            ans +=querySum(1,id[tp1], id[u]);
            //printf("ans:   %d
    ",ans);
            u = fa[tp1];
            tp1 = top[u];
        }
        //if (u == v) return ans;
        if (dep[u] > dep[v]) swap(u, v);
        ans +=querySum(1,id[u], id[v]);
        return ans;
    }
    int main() {
        scanf("%d",&n);
        for(int i=1; i<n; i++) {
            e[i].read();
            v[e[i].x].push_back(e[i].y);
            v[e[i].y].push_back(e[i].x);
        }
        for(int i=1;i<=n;i++)scanf("%d",&c[i]);
        num = 0;
        dfs1(1,0,1);
        dfs2(1,1);
        for (int i = 1; i <=n; i++) {
            val[id[i]] = c[i];
        }
        build(1,num,1);
        char s[200];
        scanf("%d",&m);
        while(m--) {
            int x,y;
            scanf("%s",&s);
            scanf("%d%d",&x,&y);
            if(s[1]=='M')
                printf("%d
    ",Qmax(x,y));
            else if (s[1] == 'H')
                update(1,id[x],y);
            else
                printf("%d
    ",Qsum(x,y));
        }
        return 0;
    }
     
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  • 原文地址:https://www.cnblogs.com/jianrenfang/p/6358540.html
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