【线段树+dfs序】J

Think:
1知识点：线段树+dfs序
2题意：输入一棵关系树，两种操作，C操作(查询结点x正在进行的工作(初始工作默认为-1))，T操作(结点x及其下属开始进行y工作)
3思路：通过dfs序将其转化为可以建为线段树的映射结点，进而进行线段树的查询和更新操作即可。（用每个节点的所包含的子节点段来当做线段树的节点，查找每个节点所包含的段可以用简单的DFS实现）
4反思：lazy标记没有很好理解，在进行更新操作时候没有考虑到应该将lazy标记down下去更新

vjudge题目链接

建议参考博客

以下为Accepted代码

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;#define MID int mid = (tree[rt].l+tree[rt].r)>>1;#define lson rt<<1#define rson rt<<1|1const int N = 51400;struct Tree{    int l, r, task;    int lazy;}tree[N<<2];int in[N], out[N], use[N];int indx;vector <int> G[N];void dfs(int k);/*dfs序*/void Build(int l, int r, int rt);void down(int rt);int Query(int p, int rt);void up_v(int L, int R, int task, int rt);int main(){    int k = 1, T, n, Q, i, u, v, task;    char st[14];    scanf("%d", &T);    while(T--){        scanf("%d", &n);        for(i = 1; i <= n; i++)            G[i].clear();/*初始化*/        memset(use, 0, sizeof(use));        for(i = 1; i <= n-1; i++){            scanf("%d %d", &u, &v);            G[v].push_back(u);/*G[v]存储v的直系下属*/            use[u] = 1;        }        indx = 0;        for(i = 1; i <= n; i++){            if(!use[i]){/*boss*/                dfs(i);                break;            }        }        Build(1, n, 1);        printf("Case #%d:\n", k++);        scanf("%d", &Q);        while(Q--){            scanf("%s", st);            if(st[0] == 'C'){                scanf("%d", &u);                printf("%d\n", Query(in[u], 1));/*in[u]为u在线段树中的位置*/            }            else if(st[0] == 'T'){                scanf("%d %d", &u, &task);                up_v(in[u], out[u], task, 1);/*区间[in[u], out[u]]为u及其下属所在线段树中的位置*/            }        }    }    return 0;}void dfs(int k){    in[k] = ++indx;    int i, len = G[k].size();    for(i = 0; i < len; i++){        dfs(G[k][i]);    }    out[k] = indx;}void Build(int l, int r, int rt){    tree[rt].lazy = 0;    tree[rt].l = l, tree[rt].r = r, tree[rt].task = -1;    if(l == r)        return ;    MID;    Build(l, mid, lson);    Build(mid+1, r, rson);}void down(int rt){    if(tree[rt].lazy){        tree[lson].lazy = tree[rson].lazy = 1;        tree[lson].task = tree[rson].task = tree[rt].task;        tree[rt].lazy = 0;    }}int Query(int p, int rt){    if(tree[rt].l == tree[rt].r)        return tree[rt].task;    down(rt);    MID;    if(p <= mid)        return Query(p, lson);    else        return Query(p, rson);}void up_v(int L, int R, int task, int rt){    if(L <= tree[rt].l && tree[rt].r <= R){        tree[rt].task = task;        tree[rt].lazy = 1;        return;    }    down(rt);/*标记更新*/    MID;    if(L <= mid)        up_v(L, R, task, lson);    if(R > mid)        up_v(L, R, task, rson);}