CF620E - New Year Tree
来源:互联网 发布:软件测试工资 编辑:程序博客网 时间:2024/06/05 23:45
题目大意:给一颗树,根节点为1,每个节点都有颜色(<=60),支持将一个子树中的所有节点都染成颜色c的操作和询问以v为根节点的子树中有多少种不同的颜色。
对子树操作,dfs序+线段树 (对树上的链操作用树链剖分)
表示颜色(col<=60) 可以用位运算(bitmasks)
两个区间颜色数的合并为 or ,总颜色统计为二进制上出现的1的个数
#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>using namespace std;#define MAXN 400100#define LL long longint n,m,l,x,y,tot,co,t,v;LL ans;int ans1;int st[MAXN],ed[MAXN],c[MAXN];int s[2*MAXN];bool vis[MAXN];struct node{ int l,r,flag; LL col;}tree[5*MAXN];struct point{ int y,next;}edge[MAXN*2];int head[MAXN];void add(int x,int y){ l++; edge[l].y=y; edge[l].next=head[x]; head[x]=l;}void dfs(int x){ st[x]=++tot; s[tot]=x; vis[x]=1; for (int i=head[x];i!=0;i=edge[i].next) if(vis[edge[i].y]==0) dfs(edge[i].y); ed[x]=++tot; s[tot]=x;}void pushup(int p){ tree[p].col=tree[p<<1].col | tree[p<<1|1].col;}void pushdown(int p){ if (tree[p].flag>0) { tree[p<<1].flag=tree[p<<1|1].flag=tree[p].flag; tree[p<<1].col=tree[p<<1|1].col=tree[p].col; tree[p].flag=0; }}void build(int p,int l,int r){ tree[p].l=l; tree[p].r=r; tree[p].flag=0; if (l==r) { tree[p].col=(LL)1 << (LL)c[s[l]]; return ; } int mid=(l+r) >> 1; build(p<<1,l,mid); build(p<<1|1,mid+1,r); pushup(p);}void change(int p,int l,int r,int c){ if (tree[p].l==l && tree[p].r==r) { tree[p].col=(LL)1 << (LL)c; tree[p].flag=c; return ; } pushdown(p); int mid=(tree[p].l+tree[p].r) >> 1; if (r<=mid) change(p<<1, l, r, c); if (l>mid) change(p<<1|1, l, r, c); if (l<=mid && r>mid) { change(p<<1,l,mid,c); change(p<<1|1,mid+1,r,c); } pushup(p);}LL query(int p,int l,int r){ if (tree[p].l==l && tree[p].r==r) { return tree[p].col; } pushdown(p); int mid=(tree[p].l+tree[p].r) >> 1; if (r<=mid) return query(p<<1,l,r); if (l>mid) return query(p<<1|1,l,r); if (l<=mid && r>mid) { LL s1=query(p<<1,l,mid); LL s2=query(p<<1|1,mid+1,r); return s1|s2; }}int main(){ scanf("%d%d", &n, &m); for (int i=1;i<=n;i++) scanf("%d", &c[i]); memset(head,0,sizeof(head)); memset(vis,0,sizeof(vis)); for (int i=1;i<n;i++) { scanf("%d%d", &x, &y); add(x,y); add(y,x); } dfs(1); build(1,1,2*n); for (int i=1;i<=m;i++) { scanf("%d%d", &t, &v); if (t==1) { scanf("%d", &co); change(1,st[v],ed[v],co); } else { ans=(LL)query(1,st[v],ed[v]); ans1=0; while (ans>0) { ans1+=ans & 1; ans>>=1; } printf("%d\n", ans1); } }}
0 0
- CF620E - New Year Tree
- Educational Codeforces Round 6 E. New Year Tree CF620E
- New Year Tree CodeForces
- New Year Tree CodeForces
- New Year Tree CodeForces
- CF 379F: New Year Tree
- Codeforces 620 E. New Year Tree
- Codeforces 620E New Year Tree
- Codeforces379F-New Year Tree(LCA)
- [CF750G]New Year and Binary Tree Paths
- new year~~
- new year
- new year
- new year
- new year
- new year
- CodeForces_GoodBye 2013_F New Year Tree LCA问题
- codeforces 379F New Year Tree 在线LCA
- 字符串替换新单词
- 线性表的顺序结构实现
- Material Design学习之 Button(详细分析,传说中的水滴动画)
- QT事件
- Linux---CentOS搭建环境之安装Tomcat
- CF620E - New Year Tree
- x265-1.7版本-common/slice.h注释
- 纪念PeakTools开工和申请博客
- Activity生命周期详解
- Python的机器学习库汇总与梳理
- ZOJ1016
- GPIO模拟IIC过程中对IIC的理解
- 关于Block的内存问题__上
- 线性表的链式结构实现