2019CCPC - Final K.Russian Dolls on the Tree

传送门：gym - 102431K

思路分析

求子树数字集合中连续段的个数
直接启发式合并就行了

样例输入

1
7
1 2
2 4
2 6
1 3
3 5
3 7

样例输出

Case #1: 1 3 3 1 1 1 1

AC代码

#include <functional>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <vector>
#include <string>
#include <cstdio>
#include <chrono>
#include <random>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#if __cplusplus >= 201103L
#include <unordered_map>
#include <unordered_set>
#endif
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0);

using namespace std;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 2e5+10;

int a[N],ans[N],vis[N],sum,flag;
vector<int>e[N];
int siz[N],son[N];

void dfs1(int u,int fa) {
    siz[u]=1;
    for(auto v:e[u]) {
        if(v==fa) continue;
        dfs1(v,u);
        siz[u]+=siz[v];
        if(siz[v]>siz[son[u]]) son[u]=v;
    }
}

void cal(int u,int fa,int keep) {
    vis[u]=keep;
    if(keep) {
        if(vis[u-1] && vis[u+1]) sum--;
        else if(!vis[u-1] && !vis[u+1]) sum++;
    } else {
        if(vis[u-1] && vis[u+1]) sum++;
        else if(!vis[u-1] && !vis[u+1]) sum--;
    }
    for(auto v:e[u]) if(v!=fa && v!=flag) cal(v,u,keep);
}

void dfs2(int u,int fa,int keep) {
    for(auto v:e[u]) if(v!=fa && v!=son[u]) dfs2(v,u,0);
    if(son[u]) {
        dfs2(son[u],u,1);
        flag=son[u];
    }
    cal(u,fa,1);
    flag=0;
    ans[u]=sum;
    if(!keep) cal(u,fa,0);
}

void solve(int time) {

    int n;
    scanf("%d",&n);
    for(int i=1; i<=n; i++) e[i].clear();
    mem(son,0);
    for(int i=1; i<n; i++) {
        int u,v;
        scanf("%d%d",&u,&v);
        e[u].push_back(v);
        e[v].push_back(u);
    }
    dfs1(1,1);
    dfs2(1,1,0);
    printf("Case #%d: ",time);
    for(int i=1; i<=n; i++) printf("%d%c",ans[i],i==n?'\n':' ');
}



int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int t;
    cin>>t;
    for(int i=1; i<=t; i++) solve(i);

    return 0;
}