Codeforces 620E New Year tree (Dfs sequence + segment tree interval operation + binary)

Test instructions: Given a tree, each node initially has a color value, with m operations, 1 operation: Input id,c. The first ID node and its subtree are dyed c.

2 operation: ID, output ID number of different colors in subtree.

Solution: First run DFS, the first ID node to control the DFS sequence is l,r interval. Then the interval operation is converted to the segment tree.

How many kinds of colors are recorded. Test instructions said the number of colors not more than 60. Then the C-color corresponds to the second-digit C-bit, and the color of the two-segment bands is the "or" of their two-segment color values. Finally, see how many 1 are in this color value binary.

#include <iostream> #include <string> #include <stdio.h> #include <string.h> #include < vector> #include <math.h> #include <queue> #include <map> #include <set> #include <
Algorithm> using namespace std;
#define MAXN 4*100005 #define LL Long Long #define INF 0x3f7f7f7f const double EPS = 1e-10;
    struct node {int l,r,c;
LL Val;
}tr[maxn<<2]; struct point {int l,r,c;}
POINT[MAXN]; struct Edge {int en,next;}
E[MAXN*2];
int p[maxn],num;
int n,m,step;
int VIS[MAXN],DFN[MAXN];
    The Dfn[i],dfs sequence is the node number that corresponds to I, void add (int st,int en) {e[num].en=en;
    E[NUM].NEXT=P[ST];
p[st]=num++;
    } void Dfs (int id) {point[id].l=++step;
    Dfn[step]=id;
    Vis[id]=1;
    int i;
        for (i=p[id];i!=-1;i=e[i].next) {int en=e[i].en;
    if (!vis[en]) dfs (EN);
    } Point[id].r=step;
cout<< "ID:" <<id<< "" <<point[id].l<< "" <<point[id].r<<endl;} void Pushup (int id) {tr[id].val=tr[id*2].val|tr[id*2+1].val;//Two period interval merged color value} void pushdown (int id) {tr[id].c=0;
    Tr[id*2].c=tr[id*2+1].c=1;
Tr[id*2].val=tr[id*2+1].val=tr[id].val;
    } void Build (int id,int l,int r) {tr[id].l=l;
    Tr[id].r=r;
    tr[id].c=0;
    if (l==r) {tr[id].val= (LL) (LL) 1<< (LL) point[dfn[l]].c);
        } else {int mid= (L+R)/2;
        Build (Id*2,l,mid);
        Build (Id*2+1,mid+1,r);
    Pushup (ID); }} void Update (int id,int l,int r,ll val) {if (l<=tr[id].l&&tr[id].r<=r) {tr[id].val= (LL) (
        (LL) 1<<val);
    Tr[id].c=1;
        } else {int mid= (TR[ID].L+TR[ID].R) >>1;
        if (TR[ID].C) pushdown (ID);
        if (r<=mid) update (id*2,l,r,val);
        else if (l>mid) update (id*2+1,l,r,val);
            else {update (id*2,l,r,val);
        Update (ID*2+1,L,R,VAL);
    } pushup (ID);
}} LL query (int id,int l,int R){if (l<=tr[id].l&&tr[id].r<=r) {return tr[id].val;
        } else {int mid= (TR[ID].L+TR[ID].R) >>1;
        if (TR[ID].C) pushdown (ID);
        LL ans;
        if (r<=mid) return query (ID*2,L,R);
        else if (l>mid) return query (ID*2+1,L,R);
            else {ans=query (id*2,l,r);
            Ans=ans|query (ID*2+1,L,R);
        return ans;
    }}} int fun (LL x) {int ans=0;
        for (ll i=0;i<=64;i++) {ll temp= (LL) (LL) 1<< (ll) i);
    if (temp&x) ans++;
} return ans;
    } int main () {int i;
        while (scanf ("%d%d", &n,&m)!=eof) {num=0;
        memset (P,-1,sizeof (p));
        for (i=1;i<=n;i++) scanf ("%d", &point[i].c);
            for (i=1;i<n;i++) {int u,v;
            scanf ("%d%d", &u,&v);
           Add (U,V);
        Add (V,u);
        } step=0; memset (vis,0,sizeof (VIS));
        DFS (1);
        Build (1,1,n);
            for (i=1;i<=m;i++) {int opt;
            int Id,co;
            scanf ("%d", &opt);
                if (opt==1) {scanf ("%d%d", &id,&co);
            Update (1,POINT[ID].L,POINT[ID].R,CO);
                } else {scanf ("%d", &id);
                LL x=query (1,POINT[ID].L,POINT[ID].R);
                cout<<i<< ":::" <<x<<endl;
            printf ("%d\n", Fun (x));
}}} return 0;
 }