HDU 6203 ping ping ping

Source：2017 ACM/ICPC Asia Regional Shenyang Online

Problem：给你一颗树，告诉你m对点是走不通的，问最少有几个坏点。

Idea：如果注意到按照lca的深度来贪心，当前u，v走得通就取那个lca，使得他的子树都无法再连出去，那么结果一定是最优的，剩下的用树状数组或者线段树维护下就好了。

Code:

#include<bits/stdc++.h>
using namespace std;

#define fi first
#define se second
#define pb push_back
#define CLR(A, X) memset(A, X, sizeof(A))
#define bitcount(X) __builtin_popcountll(X)
typedef long long LL;
typedef pair<int, int> PII;
const double eps = 1e-10;
const LL MOD = 1e9+7;
const auto INF = 0x3f3f3f3f;
int dcmp(double x) { if(fabs(x) < eps) return 0; return x<0?-1:1; }

const int MAXN = 1e4+5;

struct BIT{
int c[MAXN];
int sum(int x) { int res=0; while(x) res+=c[x], x-=(x&-x); return res; }
void add(int x, int d) { while(x<MAXN) c[x]+=d, x+=(x&-x); }
void reset() { CLR(c,0); }
}A;

struct Node {
int u, v, root, deep;
bool operator < (const Node &A) const {
return deep > A.deep;
}
};

int p[MAXN][25], deep[MAXN], dfn[MAXN], sz[MAXN], dfs_clock;
vector<int> G[MAXN];
vector<Node> g;

void dfs(int u, int fa) {
for(int i = 1; i <= 20; i++) {
p[u][i] = p[p[u][i-1]][i-1];
}
dfn[u] = ++dfs_clock, sz[u] = 1;
for(
eecd
int v:G[u]) if(v != fa) {
deep[v] = deep[u]+1;
p[v][0] = u;
dfs(v, u);
sz[u] += sz[v];
}
}

int lca(int a, int b) {
if(deep[a] < deep[b]) swap(a, b);
int t = deep[a] - deep[b];
for(int i = 0; (1<<i) <= t; i++) if((1<<i)&t) a = p[a][i];
for(int i = 20; i >= 0; i--) {
if(p[a][i] != p[b][i]) a = p[a][i], b = p[b][i];
}
return a==b?a:p[a][0];
}

int main() {
int n, u, v, m;
while(~scanf("%d", &n)) {
for(int i = 0; i <= n; i++) G[i].clear();
A.reset(), g.clear(), dfs_clock = 0, p[0][0] = 0;
for(int i = 1; i <= n; i++) {
scanf("%d%d", &u, &v);
G[u].pb(v), G[v].pb(u);
}
dfs(0, -1);
scanf("%d", &m);
while(m--) {
scanf("%d%d", &u, &v);
int root = lca(u, v);
g.pb((Node){u, v, root, deep[root]});
}
sort(g.begin(), g.end());
int ans = 0;
for(Node x:g) {
if(A.sum(dfn[x.u]) || A.sum(dfn[x.v])) continue;
A.add(dfn[x.root], 1), A.add(dfn[x.root]+sz[x.root], -1);
ans++;
}
printf("%d\n", ans);
}
return 0;
}