【CodeForces】343D Water tree （线段树好题！还未弄懂）

/*此题的方法除了用线段树求子树，通过标记父亲，更新儿子的方法，来更新祖先，学习了。对于建树的方法由于并没有说明父亲与儿子的顺序，所以需要通过两次添加。并且pre变量可以获得父亲的位置，还未弄懂！*/#define _CRT_SECURE_NO_WARNINGS#include<cstring>#include<cstdio>#include<iostream>#include<algorithm>using namespace std;#define MAX 500005#define ls rt<<1#define rs ls|1#define m (l+r)>>1int sum[MAX << 2];int col[MAX << 2];int fa[MAX << 2];int head[MAX];int num;int time;struct tree{    int v, next;}edg[MAX<<2];struct f{    int s, e;}s[MAX];void addedge(int a, int b){    edg[num].v = b;    edg[num].next = head[a];    head[a] = num++;}void dfs(int a, int pre){    fa[a] = pre;    s[a].s = ++time;    for (int i = head[a]; i != -1; i = edg[i].next)    {        if (edg[i].v == pre)continue;        dfs(edg[i].v, a);    }    s[a].e = time;}void uprt(int rt){    sum[rt] = sum[ls] && sum[rs];}void uprt2(int rt){    if (col[rt] == 1)    {        sum[rs] = sum[ls] = col[rt];        col[rs] = col[ls] = col[rt];        col[rt] = -1;    }}void updata(int L, int R, int c, int l, int r, int rt){    if (L <= l&&r <= R)    {        col[rt] = c;        sum[rt] = c;        return;    }    uprt2(rt);    int mid = m;    if (L <= mid)        updata(L, R, c, l, mid, ls);    if (mid < R)        updata(L, R, c, mid + 1, r, rs);    uprt(rt);}void updata(int q, int l, int r, int rt){    if (l == r)    {        sum[rt] = 0;        return;    }    uprt2(rt);    int mid = m;    if (q <= mid)        updata(q, l, mid, ls);    if (mid < q)        updata(q, mid + 1, r, rs);    uprt(rt);}int query(int L,int R, int l, int r, int rt){    if (L<=l&&r<=R)        return sum[rt];    uprt2(rt);    int mid = m;    int a = 0x7fff, b = 0x7fff;        if(L<=mid)a=query(L,R, l, mid, ls);        if(R>mid)b=query(L,R, mid + 1, r, rs);        return min(a, b);}int main(){    int n;    while (~scanf("%d",&n))    {        num = 0;        time = 0;        memset(head, -1, sizeof(head));        int a, b;        memset(sum, 0, sizeof(sum));        memset(col, -1, sizeof(col));        for (int i = 1; i < n; i++)        {            scanf("%d%d", &a, &b);            addedge(a, b);            addedge(b, a);        }        dfs(1, -1);        int c;        int t;        cin >> t;        bool tg = true;        while (t--)        {            scanf("%d%d", &c, &a);            if (c == 1)            {                //这个操作的作用是由于要将一个点变为1，那么假如这个点原来是0的话，那么他的祖先都是0                //但是没办法一下子全部修改祖先，所以就只能够在更新一个0的节点为1的时候将他的父亲变为0，这样就可以保持下去                //绝妙！！                if (!query(s[a].s, s[a].e, 1, n, 1) && fa[a] != -1)                {                    updata(s[fa[a]].s, 1, n, 1);                }                updata(s[a].s, s[a].e, 1, 1, n, 1);            }            else if (c == 2)            {                updata(s[a].s, 1, n, 1);            }            else if (c == 3)            {                printf("%d\n", query(s[a].s, s[a].e, 1, n, 1));            }        }    }}