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【ZOJ】 3820 Building Fire Stations

2014-10-12 20:02 337 查看
贪心可知选取的两点一定在树的直径上。。。先求出直径,然后对直径上的每个点求一下这个点不走直径能到达的最远距离,然后二分答案,判断是否可行即可。。。

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 200005
#define maxm 400005
#define eps 1e-10
#define mod 1315423911
#define INF 1e17
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
// head

struct Edge
{
int v;
Edge *next;
}pool[maxm], *H[maxn], *edges;
queue<pair<int, int> > q;
int flag[maxn], vis[maxn];
int po[maxn], most[maxn];
int dist[maxn], pre[maxn];
int n, cnt, ans1, ans2;

void init(void)
{
cnt = 0;
edges = pool;
memset(H, 0, sizeof H);
memset(flag, 0, sizeof flag);
}

void addedges(int u, int v)
{
edges->v = v;
edges->next = H[u];
H[u] = edges++;
}

void read(void)
{
int u, v;
scanf("%d", &n);
for(int i = 1; i < n; i++) {
scanf("%d%d", &u, &v);
addedges(u, v);
addedges(v, u);
}
}

int bfs(int s)
{
int mx = 0, t = s;
memset(vis, 0, sizeof vis);
dist[s] = 0, vis[s] = 1, pre[s] = 0;
q.push(mp(s, 0));
while(!q.empty()) {
pair<int, int> pp = q.front();
q.pop();
int u = pp.first, d = pp.second;
if(dist[u] > mx) mx = dist[u], t = u;
for(Edge *e = H[u]; e; e = e->next) {
int v = e->v;
if(!vis[v]) {
vis[v] = 1;
dist[v] = d + 1;
pre[v] = u;
q.push(mp(v, dist[v]));
}
}
}
return t;
}

void find(int now)
{
while(now) {
po[++cnt] = now;
flag[now] = 1;
now = pre[now];
}
}

int BFS(int s)
{
int mx = 0;
vis[s] = 1, q.push(mp(s, 0));
while(!q.empty()) {
pair<int, int> pp = q.front();
q.pop();
int u = pp.first, d = pp.second;
mx = max(mx, d);
for(Edge *e = H[u]; e; e = e->next) {
int v = e->v;
if(!vis[v] && !flag[v]) {
vis[v] = 1;
q.push(mp(v, d+1));
}
}
}
return mx;
}

bool check(int d)
{
int a = 0, b = cnt+1;
int tt = 0;
for(int i = 1; i <= cnt; i++) {
if(tt > d || most[i] > d) break;
else a = i, ans1 = po[i], tt += 1;
}
tt = 0;
for(int i = cnt; i >= 1; i--) {
if(tt > d || most[i] > d) break;
else b = i, ans2 = po[i], tt += 1;
}
if(ans1 == ans2) ans2 = po[b+1];
//	printf("a = %d b = %d\n", a, b);
if(a >= b) return true;
int ok = 1;
for(int i = a; i <= b; i++) if(min(i - a + most[i], b - i + most[i]) > d) ok = 0;
return ok;
}

void work(void)
{
int s1, s2;
s1 = bfs(1);
s2 = bfs(s1);
find(s2);
//	printf("s1 = %d s2 = %d\n", s1, s2);
//	for(int i = 1; i <= cnt; i++) printf("%d\n", po[i]);
memset(vis, 0, sizeof vis);
for(int i = 1; i <= cnt; i++) most[i] = BFS(po[i]);

//	printf("AA %d BB\n", check(2));

int bot = 0, top = dist[s2], mid, res;
while(top >= bot) {
mid = (top + bot) >> 1;
if(check(mid)) res = mid, top = mid - 1;
else bot = mid + 1;
}
check(res);
printf("%d %d %d\n", res, ans1, ans2);
}

int main(void)
{
int _;
while(scanf("%d", &_)!=EOF) {
while(_--) {
init();
read();
work();
}
}

return 0;
}
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