Codeforces 343D Water Tree dfs序+线段树
题目链接：http://codeforces.com/contest/343/problem/D
题意：给定一棵树（给定图一定是树）
1、把v点及其子树灌上水
2、把v点及v到根的路径去掉水
3、询问v点是否有水
思路：
dfs序（不是欧拉序列）把树转成dfs_clock
那么对于点v 出现的时间in[v]和消失的时间out[v] ，一定会把v子树下所有节点都夹在[ in[v], out[v] ]之中，则操作1就是把 [in[v], out[v]]改成1
把路径去掉水，显然是不存在这样直接到达根部的链，所以单点更新 in[v] = 0，若询问[in[u], out[u]]时 区间内存在一个0，则u子树下存在0，即u是没有水的。
注意的是，此时如果把u这个区间灌上水，则会把v这个去水效果去掉，即在线段树上找不到痕迹，所以把 Father[u] 这个点单点更新成0。
AC1:
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<vector>
#include<set>
using namespace std;
#define N 510000
#define L(x) (x<<1)
#define R(x) (x<<1|1)
inline int Mid(int a,int b){return (a+b)>>1;}
struct Edge{
int from, to, nex;
}edge[N<<1];
int head[N], edgenum;
void add(int u,int v){
Edge E={u,v,head[u]};
edge[edgenum] = E;
head[u] = edgenum++;
}
struct node{
int l,r,minn;
}tree[N<<2];
void push_up(int id){
tree[id].minn = min(tree[id].minn,min(tree[L(id)].minn,tree[R(id)].minn));
}
void push_down(int id){
if(tree[id].l==tree[id].r)return;
if(tree[id].minn){
tree[L(id)].minn = tree[R(id)].minn = 1;
}
}
void build(int l,int r,int id){
tree[id].l = l, tree[id].r = r;
tree[id].minn = 0;
if(l==r)return ;
int mid = Mid(l,r);
build(l,mid,L(id)); build(mid+1,r,R(id));
}
void updata_point(int pos,int id){
push_down(id);
if(tree[id].l==tree[id].r){
tree[id].minn = 0;
return;
}
int mid = Mid(tree[id].l,tree[id].r);
if(pos<=mid)
updata_point(pos,L(id));
else
updata_point(pos,R(id));
push_up(id);
}
void updata_inter(int l,int r,int id){
push_down(id);
if(l == tree[id].l && tree[id].r == r){
tree[id].minn = 1;
return ;
}
int mid = Mid(tree[id].l, tree[id].r);
if(mid<l)
updata_inter(l,r,R(id));
else if(r<=mid)
updata_inter(l,r,L(id));
else {
updata_inter(l,mid,L(id));
updata_inter(mid+1,r,R(id));
}
push_up(id);
}
int query(int l,int r,int id){
push_down(id);
if(l==tree[id].l && tree[id].r == r)
return tree[id].minn;
int mid = Mid(tree[id].l,tree[id].r);
if(mid<l)
return query(l,r,R(id));
else if(r<=mid)
return query(l,r,L(id));
else return min(query(l,mid,L(id)),query(mid+1,r,R(id)));
}
int n;
int in[N], out[N], Time, Father[N];
void dfs(int u,int fa){
Father[u] = fa;
in[u] = ++Time;
for(int i = head[u]; ~i; i = edge[i].nex){
int v = edge[i].to; if(v==fa)continue;
dfs(v,u);
}
out[u] = Time;
}
void init(){
memset(head, -1, sizeof head);edgenum=0;
}
int main(){
int i, j, u, v, que;
while(~scanf("%d",&n)){
init();
for(i=1;i<n;i++){
scanf("%d %d",&u,&v);
add(u,v); add(v,u);
}
Time = 0;
dfs(1,-1);
build(1,Time,1);
scanf("%d",&que);
while(que--){
scanf("%d %d",&u,&v);
if(u==1){
int tmp = query(in[v],out[v],1);
updata_inter(in[v],out[v],1);
if(Father[v]!=-1 && tmp==0)
updata_point(in[Father[v]],1);
}
else if(u==2){
updata_point(in[v],1);
}
else
printf("%d\n",query(in[v],out[v],1));
}
}
return 0;
}
/*
5
1 2
5 1
4 2
2 3
12
1 1
2 3
3 1
3 2
3 3
3 4
1 2
2 4
3 1
3 3
3 4
3 5
*/
AC2：
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<vector>
#include<set>
using namespace std;
#define N 1010000
#define L(x) (x<<1)
#define R(x) (x<<1|1)
inline int Mid(int a,int b){return (a+b)>>1;}
struct Edge{
int from, to, nex;
}edge[N<<1];
int head[N], edgenum;
void add(int u,int v){
Edge E={u,v,head[u]};
edge[edgenum] = E;
head[u] = edgenum++;
}
struct node{
int l,r,sum,lazy;
int size(){return r-l+1;}
}tree[N<<2];
void push_up(int id){
if(tree[id].l==tree[id].r)return;
tree[id].sum = tree[L(id)].sum + tree[R(id)].sum;
}
void push_down(int id){
if(tree[id].l==tree[id].r)return;
if(tree[id].lazy){
if(tree[id].sum)
{
tree[L(id)].sum = tree[L(id)].size();
tree[R(id)].sum = tree[R(id)].size();
}
else
{
tree[L(id)].sum = tree[R(id)].sum = 0;
}
tree[L(id)].lazy = tree[R(id)].lazy = 1;
tree[id].lazy = 0;
}
}
void build(int l,int r,int id){
tree[id].l = l, tree[id].r = r;
tree[id].sum = tree[id].lazy = 0;
if(l==r)return ;
int mid = Mid(l,r);
build(l,mid,L(id)); build(mid+1,r,R(id));
}
void updata(int l,int r,int val,int id){
push_down(id);
if(l == tree[id].l && tree[id].r == r){
if(val)
{
tree[id].sum = tree[id].size();
tree[id].lazy = 1;
}
else
{
tree[id].sum = 0;
tree[id].lazy = 0;
}
return ;
}
int mid = Mid(tree[id].l, tree[id].r);
if(mid<l)
updata(l,r,val,R(id));
else if(r<=mid)
updata(l,r,val,L(id));
else {
updata(l,mid,val,L(id));
updata(mid+1,r,val,R(id));
}
push_up(id);
}
int query(int l,int r,int id){
push_down(id);
if(l==tree[id].l && tree[id].r == r)
return tree[id].sum;
int mid = Mid(tree[id].l,tree[id].r);
if(mid<l)
return query(l,r,R(id));
else if(r<=mid)
return query(l,r,L(id));
else return query(l,mid,L(id)) + query(mid+1,r,R(id));
}
int n;
int in[N], out[N], Time, si[N], pa[N];
void dfs(int u,int fa){
in[u] = ++Time;
si[u] = 1;
pa[u] = fa;
for(int i = head[u]; ~i; i = edge[i].nex){
int v = edge[i].to; if(v==fa)continue;
dfs(v,u);
si[u] += si[v];
}
out[u] = Time;
}
void init(){
memset(head, -1, sizeof head);edgenum=0;
}
int main(){
int i, j, u, v, que;
while(~scanf("%d",&n)){
init();
for(i=1;i<n;i++){
scanf("%d %d",&u,&v);
add(u,v); add(v,u);
}
Time = 0;
dfs(1,-1);
build(1,Time,1);
scanf("%d",&que);
while(que--){
scanf("%d %d",&u,&v);
if(u==1){
int tmp = query(in[v],out[v],1);
updata(in[v],out[v],1,1);
if(tmp!=si[v] && pa[v]!=-1)
updata(in[pa[v]],in[pa[v]],0,1);
}
else if(u==2){
updata(in[v],in[v],0,1);
}
else
printf("%d\n",query(in[v],out[v],1)==si[v]);
}
}
return 0;
}
/*
5
1 2
5 1
4 2
2 3
12
1 1
2 3
3 1
3 2
3 3
3 4
1 2
2 4
3 1
3 3
3 4
3 5
*/