Inzane finally found Zane with a lot of money to spare, so they together decided to establish a country of their own.

Ruling a country is not an easy job. Thieves and terrorists are always ready to ruin the country's peace. To fight back, Zane and Inzane have enacted a very effective law: from each city it must be possible to reach a police station by traveling at most d kilometers along the roads.

There are n cities in the country, numbered from 1 to n, connected only by exactly n - 1 roads. All roads are 1 kilometer long. It is initially possible to travel from a city to any other city using these roads. The country also has k police stations located in some cities. In particular, the city's structure satisfies the requirement enforced by the previously mentioned law. Also note that there can be multiple police stations in one city.

However, Zane feels like having as many as n - 1 roads is unnecessary. The country is having financial issues, so it wants to minimize the road maintenance cost by shutting down as many roads as possible.

Help Zane find the maximum number of roads that can be shut down without breaking the law. Also, help him determine such roads.

Input

The first line contains three integers n, k, and d (2 ≤ n ≤ 3·105, 1 ≤ k ≤ 3·105, 0 ≤ d ≤ n - 1) — the number of cities, the number of police stations, and the distance limitation in kilometers, respectively.

The second line contains k integers p1, p2, ..., pk (1 ≤ pi ≤ n) — each denoting the city each police station is located in.

The i-th of the following n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the cities directly connected by the road with index i.

It is guaranteed that it is possible to travel from one city to any other city using only the roads. Also, it is possible from any city to reach a police station within d kilometers.

Output

In the first line, print one integer s that denotes the maximum number of roads that can be shut down.

In the second line, print s distinct integers, the indices of such roads, in any order.

If there are multiple answers, print any of them.

Examples
Input

Copy
6 2 4
1 6
1 2
2 3
3 4
4 5
5 6
Output

Copy
1
5
Input

Copy
6 3 2
1 5 6
1 2
1 3
1 4
1 5
5 6
Output

Copy
2
4 5
Note

In the first sample, if you shut down road 5, all cities can still reach a police station within k = 4 kilometers.

In the second sample, although this is the only largest valid set of roads that can be shut down, you can print either 4 5 or 5 4 in the second line.

将每个警察局压入queue中,然后 bfs搜索;

如果某个点的 to 已经被访问了,但该条边还没有被访问,ans++;

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
} ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; } /*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/ int n, k, d;
struct node {
int to;
int id;
node(int to,int id):to(to),id(id){}
node(){}
}edge[maxn];
queue<int>q;
int ans;
int vis[maxn];
vector<node>vc[maxn];
int fgedge[maxn]; void bfs() {
while (!q.empty()) {
int u = q.front(); q.pop();
for (int i = 0; i < vc[u].size(); i++) {
node tmp = vc[u][i];
if (fgedge[tmp.id])continue;
if (vis[tmp.to]) {
ans++;
fgedge[tmp.id] = 2;
continue;
}
vis[tmp.to] = 1; fgedge[tmp.id] = 1;
q.push(tmp.to);
}
}
} int main() {
//ios::sync_with_stdio(0);
cin >> n >> k >> d;
for (int i = 1; i <= k; i++) {
int x; rdint(x); vis[x] = 1;
q.push(x);
}
for (int i = 1; i < n; i++) {
int u, v; rdint(u); rdint(v);
vc[u].push_back(node(v, i));
vc[v].push_back(node(u, i));
}
bfs();
cout << ans << endl;
for (int i = 1; i < n; i++) {
if (fgedge[i] == 2)cout << i << ' ';
} return 0;
}

CF796D Police Stations 思维的更多相关文章

  1. CF796D Police Stations BFS+染色

    题意:给定一棵树,树上有一些点是警察局,要求所有点到最近的警察局的距离不大于 $d$,求最多能删几条边 ? 题解: 考虑什么时候一条边可以被断开:这条边的两个端点被两个不同的警察局覆盖掉. 我们要设计 ...

  2. 96D - Police Stations

    96D - Police Stations 思路:bfs,从所有的警察局开始bfs,因为bfs的深度一样,而且题目给的树保证满足条件,所以不用考虑深度. 如果搜索到一个点a,他的下一个点b已经被搜索过 ...

  3. Codeforces Round #408 (Div. 2) D - Police Stations

    地址:http://codeforces.com/contest/796/problem/D 题目: D. Police Stations time limit per test 2 seconds ...

  4. CodeForces - 796D Police Stations bfs

    思路:删除尽量多的边使得所有点都能在限制距离之内到达一个警局,删除边会形成多棵子树,最多只能k棵.其实就是以每个警局为根结点,把整棵树划分为以警局为根结点的k棵树,说明要删除的边的数量就是k-1条,即 ...

  5. Police Stations CodeForces - 796D (bfs)

    大意: 给定树, 有k个黑点, 初始满足条件:所有点到最近黑点距离不超过d, 求最多删除多少条边后, 使得原图仍满足条件. 所有黑点开始bfs, 贪心删边. #include <iostream ...

  6. Codeforces Round #408 (Div. 2) D. Police Stations(最小生成树+构造)

    传送门 题意 n个点有n-1条边相连,其中有k个特殊点,要求: 删去尽可能多的边使得剩余的点距特殊点的距离不超过d 输出删去的边数和index 分析 比赛的时候想不清楚,看了别人的题解 一道将1个联通 ...

  7. 【codeforces 796D】Police Stations

    [题目链接]:http://codeforces.com/contest/796/problem/D [题意] 在一棵树上,保证每个点在距离d之内都有一个警察局; 让你删掉最多的边,使得剩下的森林仍然 ...

  8. ZOJ 2699 Police Cities

    Police Cities Time Limit: 10 Seconds      Memory Limit: 32768 KB Once upon the time there lived a ki ...

  9. Codeforces Round #408 (Div. 2)

    C. Bank Hacking 题目大意:给出一棵n个节点的树,每个节点有一个权值,删掉一个点的代价为当前这个点的权值,并且会使其相邻点和距离为2且中间隔着未被删除的点的点权值加1,现在选一个点开始删 ...

随机推荐

  1. Java 的三个注释

    单行注释 // 这是名为 a 的类 class a{ } 多行注释 /* 这是多行注释 可以注释多行 */ class a{ } 文档注释 /** 这是文档注释 可以注释多行 */ class a{ ...

  2. Eclipse与github整合完整版

    最近朋友都推荐使用github管理自己的项目,而且免费用户可以有5个仓库,恰好我也想了解下git,借此机会学习一下. github官方指南使用独立第三方git工具来进行版本控制,并不借助于eclips ...

  3. Enumeration与Iterator的对比

    Enumeration与Iterator的对比 Enumeration 接口 Iterator 接口 参数的含义 枚举类型 迭代器元素类型 所在包 java.util 父类 无 子类 StringTo ...

  4. 4-4 zk特性 – 理解watcher机制

    watcher是zk里面非常重要的特性.watcher一定要去好好地看一下,一定要去好好地理解一下它是如何去用的,包括触发的事件类型等等.监督者也可以理解为触发器,也就是说当我们的节点发生了一些变化的 ...

  5. c语言学习笔记 const变量

    在c语言的编程过程中经常会遇到有常数参加运算的运算,比如这种. int a=100*b; 这个100我们叫常数或者叫常量,但是程序中我们不推荐这种直接写常数的方法,有两个缺点. 第一是程序可读性差. ...

  6. Android 中的菜单 OptionsMenu的简单应用

    OptionsMenu就是安卓手机中的菜单选项 首先 要实现对菜单的操作就要先重写OnCreateOptionsMenu(Menu menu)方法 通常有两种方法来实现对菜单中选项的添加 第一种是动态 ...

  7. python3-file的修改实现类似shell中sed的功能

    # Auther: Aaron Fan '''思路:目的是为了修改yesterday这个文件,但是因为无法直接去修改这个文件,所以需要先把修改好的内容写入高yesterday.new这个文件中,然后再 ...

  8. Django rest-framework框架十大功能分析

    rest-framework框架有哪些作用? 一共有十点. 路由 - 可以通过as_view传参数,根据请求方式不同执行相应的方法 - 可以在url中设置一个结尾,类似于: .json 视图 - 帮助 ...

  9. 本地方法中printf如何传给java--java系统级命名管道

    本地方法中printf如何传给java--java系统级命名管道 摘自:https://blog.csdn.net/dog250/article/details/6007301 2010年11月13日 ...

  10. 查看类属性和方法---structure