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，即删除的边的条数是一定的。剩下就是为每个节点找根结点，考虑从所有警局出发得到到每个点的最短距离，则当前节点u，一定是从u->v，如果d[v] <= lim则这条边一定会保留。
AC代码
#include <cstdio>
#include <cmath>
#include <cctype>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = 3e5 + 5;
int vis[maxn], d[maxn];
int n, k, lim;
map<PI, int>ha;
vector<int>G[maxn];
queue<int>q;
int main() {
while(scanf("%d%d%d", &n, &k, &lim) == 3) {
ha.clear();
while(!q.empty()) q.pop();
for(int i = 1; i <= n; ++i) G[i].clear();
memset(d, -1, sizeof(d));
int pos;
for(int i = 0; i < k; ++i) {
scanf("%d", &pos);
d[pos] = 0;
q.push(pos);
}
int u, v;
for(int i = 0; i < n-1; ++i) {
scanf("%d%d", &u, &v);
ha[make_pair(u, v)] = i+1;
ha[make_pair(v, u)] = i+1;
G[u].push_back(v);
G[v].push_back(u);
}
memset(vis, 0, sizeof(vis));
int cnt = 0; //要保留的边的数量
while(!q.empty()) {
int u = q.front(); q.pop();
for(int i = 0; i < G[u].size(); ++i) {
int v = G[u][i];
if(d[v] == -1) {
d[v] = d[u] + 1;
q.push(v);
if(d[v] <= lim) {
int id = ha[make_pair(u, v)];
vis[id] = 1;
++cnt;
}
}
}
}
printf("%d\n", n-cnt-1);
//printf("%d\n", k-1);
for(int i = 1; i <= n-1; ++i) {
if(!vis[i]) printf("%d ", i);
}
printf("\n");
}
return 0;
} 如有不当之处欢迎指出！