Assign the task HDU3974

题目链接

题目大意是给一个固定结构的树（结点个数为N , N<=5e4），现在M(M<=5e4)组修改或查询，修改为 x , y 将结点x及其所有后代结点染成颜色y，查询为 x 询问 结点x当前的颜色。

其实就是区间染色问题，不过需要预处理DFS两次，第一次DFS预处理每个结点的后代个数，第二次DFS根据每个结点的后代个数为当前结点分配区间。然后按染色问题处理。

#include <bits/stdc++.h>
#define Lson(x) ((x)<<1)
#define Rson(x) ((x)<<1|1)
using namespace std;

typedef pair<int,int> pii;
const pii leisure = make_pair(0,-1);
const int maxn = 5e4;
struct Edge{
    int v;
    int next;
}edges[maxn+10];
int tot = 0;
int head[maxn + 10];
bool indegree[maxn+10];
int cntsons[maxn + 10];
pii d[(maxn<<2)+10];
pii subordinates[maxn + 10];

void addedge(int u,int v){
    edges[tot].v = v;
    edges[tot].next = head[u];
    head[u] = tot ++;
}

void preDFS(int cur){
    cntsons[cur] = 0;
    for(int i = head[cur]; i!=-1; i = edges[i].next){
        int son = edges[i].v;
        preDFS(son);
        cntsons[cur] += 1 + cntsons[son];
    }
}

void DFS(int cur,int L,int R){
    subordinates[cur] = make_pair(L,R);
    int LL = L + 1;
    for(int i=head[cur];i!=-1;i = edges[i].next){
        int v = edges[i].v;
        DFS(v,LL,LL + cntsons[v]);
        LL = LL + cntsons[v] + 1;
    }
}

void build(int s,int t,int p){
    d[p] = leisure;
    if(s == t){
        return;
    }
    int mid = (s+t) >> 1;
    build(s,mid,Lson(p));
    build(mid+1,t,Rson(p));
}

void update(int L,int R,pii col,int s,int t,int p){
    if(L == s && t == R){
        d[p] = max(d[p],col);
        return;
    }
    int mid = (s + t) >> 1;
    if(R <=mid) update(L,R,col,s,mid,Lson(p));
    else if( L > mid) update(L,R,col,mid+1,t,Rson(p));
    else update(L,mid,col,s,mid,Lson(p)) , update(mid+1,R,col,mid+1,t,Rson(p));
}

int query(int L,pii ever,int s,int t,int p){
    ever = max(ever,d[p]);
    if(s==t){
        return ever.second;
    }
    int mid = (s + t) >> 1;
    if(L <= mid) return query(L,ever,s,mid,Lson(p));
    else query(L,ever,mid+1,t,Rson(p));
}

int main(){
    int T;
    scanf("%d",&T);
    for(int cntT=1;cntT<=T;cntT++){
        printf("Case #%d:
",cntT);
        int N;
        scanf("%d",&N);
        tot = 0;
        for(int i = 1;i<=N;++i) {
            head[i] = -1;
            indegree[i] = 0;
        }
        for(int i=0;i<N-1;++i){
            int v,u;
            scanf("%d%d",&v,&u);
            addedge(u,v);
            indegree[v] = true;
        }
        int root = 1;
        for(int i=1;i<=N;++i){
            if(!indegree[i]){
                root = i;
                preDFS(i);
                break;
            }
        }
        DFS(root,1,N);
        build(1,N,1);
//        for(int i=1;i<=N;++i){
//            printf("%d : %d , %d
",i,subordinates[i].first , subordinates[i].second);
//        }
        int M;
        scanf("%d",&M);
        int choke = 0;
        for (int i = 0; i < M; ++i) {
            char op[3];
            scanf("%s",op);
            if(op[0] == 'C'){
                int x;
                scanf("%d",&x);
                int L = subordinates[x].first ;
                printf("%d
",query(L,leisure,1,N,1));
            }else{
                int x,y;
                scanf("%d%d",&x,&y);
                int L = subordinates[x].first , R = subordinates[x].second;
                pii workcol = make_pair(++choke , y);
                update(L,R,workcol,1,N,1);
            }
        }
    }
    return 0;
}