Codeforces Gym 100650B Countdown （离线）

题目链接：http://codeforces.com/gym/100650

根据给出的树和d，求出一些结点，这些结点形成子树的第d层结点数应该尽量多，具体要求可以参考题目。

dfs一个结点前保存询问深度的答案，访问完以后减去之前的值就得到答案了。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1100;

vector<string> names;
map<string,int> mp;
int id_cnt;
#define se second
#define MP make_pair

int ID(const string & s){
map<string,int>::iterator i = mp.find(s);
if(i != mp.end()) return i->se;
names.push_back(s);
mp.insert(MP(s,id_cnt));
return id_cnt++;
}
int deg[maxn];
int head[maxn],nxt[maxn],to[maxn],ecnt;

void addEdge(int u,int v)
{
to[ecnt] = v;
nxt[ecnt] = head[u];
head[u] = ecnt++;
}

int pre[maxn*2];
int cnt[maxn*2];
int deep[maxn];

struct Node
{
int d,id;
bool operator < (const Node & x) const {
return d < x.d || ( d == x.d && names[id] > names[x.id] );
}
};

priority_queue<Node> q;

int n,d;
void dfs(int u)
{
cnt[deep[u]]++;
int qd = deep[u]+d;
if(qd < maxn)   pre[u] =  cnt[qd];
for(int i = head[u]; ~i; i = nxt[i]){
int v = to[i];
deep[v] = deep[u]+1;
dfs(v);
}
if(qd<maxn){
int ans = cnt[qd]-pre[u];
if(ans) q.push(Node{ans,u});
}
}

void init()
{
names.clear();
mp.clear();
id_cnt = 0;
ecnt = 0;
memset(deg,0,sizeof(deg));
memset(head,-1,sizeof(head));
memset(cnt,0,sizeof(cnt));
while(q.size()) q.pop();
}

char str[500];

void read()
{
scanf("%d%d",&n,&d);
while(n--){
int ch; scanf("%s%d",str,&ch);
int fa = ID(str);
while(ch--){
scanf("%s",str);
int v = ID(str);
deg[v]++;
addEdge(fa,v);
}
}
}

void topo()
{
int root;
n = mp.size();
for(int i = 0; i < n; i++){
if(deg[i] == 0) { root = i; break; }
}
deep[root] = 0;
dfs(root);
}

int main()
{
//freopen("in.txt","r",stdin);
int T; scanf("%d",&T);
for(int kas = 1; kas <= T; kas++){
init();
read();
topo();
printf("Tree %d:\n",kas);
for(int i = 1; i <= 3; i++){
if(q.empty()) break;
Node x = q.top(); q.pop();
printf("%s %d\n",names[x.id].c_str(),x.d);
if(i == 3){
int pre = x.d;
while(q.size()){
x = q.top(); q.pop();
if(x.d == pre){
printf("%s %d\n",names[x.id].c_str(),x.d);
}else {
break;
}
}
}
}
if(kas != T) putchar('\n');
}
return 0;
}