HDU4358【離散化+DFS序+莫隊演算法】
阿新 • • 發佈:2019-02-08
思路:
這個DFS序還是蠻有意思的~然後就可以把以u為根的樹的狀態直接轉化到區間(一維陣列)上,而且陣列是不大的。
然後就變成了區間處理,套一下莫隊就好了。
Code:
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define mem(a, b) memset(a, b, sizeof(a))
const int Maxn = 1e5 + 10;
struct Edge{int v, nex;}edge[Maxn<<1];
int head[Maxn], tol;
void init(){tol = 0,mem(head, -1);}
void addedge(int u, int v){
edge[tol] = (Edge){v, head[u]}, head[u] = tol++;
edge[tol] = (Edge){u, head[v]}, head[v] = tol++;
}
int w[Maxn], _time, val[Maxn], n, k, q, a[Maxn];
int Left[Maxn], Right[Maxn];
vector<int>xs;
struct asd{
int id;
int Left, Right;
}node[Maxn];
int res[Maxn], ans, sum[Maxn];
int pos[Maxn];
bool cmp(asd x,asd y)
{
if(pos[x.Left] == pos[y.Left])
return x.Right<y.Right;
return pos[x.Left]<pos[y.Left];
}
void DFS(int u, int fa){
int v;
Left[u] = ++_time, w[_time] = a[u];
for(int i=head[u]; ~i;i = edge[i].nex){
v = edge[i].v;
if (v == fa) continue;
DFS(v, u);
}
Right[u] = _time;
}
void ADD(int ww){
if(sum[ww] == k) ans--;
sum[ww]++;
if(sum[ww] == k) ans++;
}
void DEL(int ww){
if(sum[ww] == k) ans--;
sum[ww]--;
if(sum[ww] == k) ans++;
}
void solve(){
ans=0;
mem(sum, 0);
sort(node+1, node+1+q, cmp);
for(int i=node[1].Left;i<=node[1].Right;i++){
ADD(w[i]);
}
res[node[1].id]=ans;
int L=node[1].Left,R=node[1].Right;
for(int i=2; i<=q; i++){
while(R<node[i].Right){
R++;
ADD(w[R]);
}
while(R>node[i].Right){
DEL(w[R]);
R--;
}
while(L<node[i].Left){
DEL(w[L]);
L++;
}
while(L>node[i].Left){
L--;
ADD(w[L]);
}
res[node[i].id]=ans;
}
for(int i=1;i<=q;i++)
printf("%d\n", res[i]);
}
int main()
{
int T, u, v, L, R;
scanf("%d", &T);
for(int cas = 1; cas <= T; cas++){
scanf("%d%d",&n, &k);
xs.clear();
int block=(int)sqrt(n);
for(int i=1;i<=n;i++){
scanf("%d", &a[i]);
xs.push_back(a[i]);
pos[i]=(i-1)/block+1;
}
sort(xs.begin(), xs.end());
xs.erase(unique(xs.begin(), xs.end()), xs.end());
for(int i=1;i<=n;++i) a[i] = lower_bound(xs.begin(), xs.end(), a[i]) - xs.begin() + 1;
init();
for(int i=1;i<n;i++){
scanf("%d%d", &u, &v);
addedge(u, v);
}
_time = 0;
DFS(1, -1);
scanf("%d", &q);
for(int i=1;i<=q;i++){
scanf("%d", &u);
node[i].id = i;
node[i].Left = Left[u], node[i].Right = Right[u];
}
printf("Case #%d:\n", cas);
solve();
if(cas != T) puts("");
}
return 0;
}