ZOJ 3820 2014ACM/ICPC牡丹江赛区B题

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3797714 2014 - 10 - 12 21:58 : 19 Accepted 3820 C++ 1350 70240 zz_1215
比较麻烦的一道题吧，开始的时候不停的段异常，后面知道是爆栈了，然后用数组模拟递归，才ac了

思路挺简单的，先找到这个树的直径，单独拿出来，可以证明最后选的两个点一定是在直径上的，我就不证了

然后求出这条直径上的每个点向外延伸的最远距离

对这个距离做两次RMQ，第一次是对于往左边计算最大距离，所以要这个距离的序列要依次+1，+2，+3.......+n-1，+n

第二次是对于往右边计算最大距离，所以序列要+n，+n-1........+3，+2，+1

然后确定一个左起点，二分右起点

最后的复杂度是O（n*log（n））

实现起来还是挺复杂的，吐槽下这种考代码能力的题

最后有一点提醒下，向外延伸的时候不要忘了向父节点的方向延伸，因为这个我又wa了一次

#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;
#ifdef _WIN32
#define i64 __int64
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
#define i64 long long
#define out64 "%lld\n"
#define in64 "%lld"
#endif
/************ for topcoder by zz1215 *******************/
#define foreach(c,itr)  for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FF(i,a)         for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a,b)      for( int i = (a) ; i >= (b) ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define pb              push_back
#define pf              push_front
#define X               first
#define Y               second
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define MC(a,b)		memcpy(a,b,sizeof(b))
#define MAX(a,b)        ((a)>(b)?(a):(b))
#define MIN(a,b)        ((a)<(b)?(a):(b))
#define read            freopen("out.txt","r",stdin)
#define write           freopen("out2.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-9;
const double pi = acos(-1.0);
const int maxn = 211111;

struct Node{
	int now;
	int to;
	int h;
	bool operator < (const Node & cmp) const{
		return h>cmp.h;
	}
}node;

int n;
vector<Node>g[maxn];
int dfv[maxn];
int dfn[maxn];
int t[maxn];
int h[maxn];

int df;
vector<int>s;
vector<int>si;
bool vis[maxn];
vector<int>line;
int a[maxn];
int ax[maxn][20];
int dx[maxn][20];
int lg2[maxn];
bool isline[maxn];

void dfs(){
	for (int i = 1; i <= n; i++){
		vis[i] = false;
	}
	df = 1;
	s.clear();
	si.clear();
	s.push_back(1);
	si.push_back(0);
	vis[1] = true;
	int now, to, id;
	while (!s.empty()){
		now = s.back();
		id = si.back();
		if (id < g[now].size()){
			to = g[now][id].to;
			si.back()++;
			if (!vis[to]){
				vis[to] = true;
				t[to] = now;
				s.push_back(to);
				si.push_back(0);
			}
		}
		else{
			dfv[df] = s.back();
			dfn[s.back()] = df++;
			s.pop_back();
			si.pop_back();
		}
	}
}

int find_len(int now){
	if (g[now].size() >= 2){
		return g[now][0].h + g[now][1].h;
	}
	else if (g[now].size() == 1){
		return g[now][0].h;
	}
	else{
		return 0;
	}
}

void get_line(int now){
	if (g[now].size() >= 2){
		int t1 = g[now][0].to;
		int t2 = g[now][1].to;
		isline[now] = true;
		while (true){
			isline[t1] = true;
			line.push_back(t1);
			if (g[t1].size() > 0){
				t1 = g[t1][0].to;
			}
			else{
				break;
			}
		}
		
		reverse(line.begin(), line.end());
		line.push_back(now);
		while (true){
			isline[t2] = true;
			line.push_back(t2);
			if (g[t2].size() > 0){
				t2 = g[t2][0].to;
			}
			else{
				break;
			}
		}
	}	
	else if(g[now].size() ==1){
		while (true){
			isline[now] = true;
			line.push_back(now);
			if (g[now].size() > 0){
				now = g[now][0].to;
			}
			else{
				break;
			}
		}
	}
}

int max_way(int now){
	int to;
	int re = 0;
	for (int i = 0; i < g[now].size(); i++){
		to = g[now][i].to;
		if (!isline[to]){
			re = max(re, g[now][i].h);
		}
	}	
	return re;
}

void sparse_table(){
	for (int i = 0; i < line.size(); i++){
		ax[i][0] = a[i]+i;
		dx[i][0] = a[i]+(int)line.size()-1-i;
	}
	for (int step = 1; (1 << step) < line.size(); step++){
		for (int i = 0; i < line.size(); i++){
			ax[i][step] = ax[i][step - 1];
			dx[i][step] = dx[i][step - 1];
			if (i + (1 << (step - 1)) < line.size()){
				ax[i][step] = max(ax[i][step], ax[i + (1 << (step - 1))][step - 1]);
				dx[i][step] = max(dx[i][step], dx[i + (1 << (step - 1))][step - 1]);
			}
		}
	}
}

int max_a(int l, int r){
	return max(ax[l][lg2[r - l + 1]], ax[r - (1 << lg2[r - l + 1]) + 1][lg2[r - l + 1]]);
}

int max_d(int l, int r){
	return max(dx[l][lg2[r - l + 1]], dx[r - (1 << lg2[r - l + 1]) + 1][lg2[r - l + 1]]);
}

int find(int l, int r){
	int mid = (r + l) / 2;
	return max(max_a(l, mid) - l, max_d(mid+1,r)-( (int)line.size()-1-r ) );
}

int back[maxn];

void find_back(){
	for (int i = 1; i <= n; i++){
		back[i] = 0;
	}
	queue<int>q;
	q.push(1);
	int now, to,fa,temp;
	while (!q.empty()){
		now = q.front();
		q.pop();
		if (t[now]){
			fa = t[now];
			back[now] = 1 + back[fa];
			temp = 0;
			if (g[fa][0].to == now){
				if (g[fa].size() >= 2){
					temp = g[fa][1].h+1;
				}
			}
			else{
				temp = g[fa][0].h;	
			}
			back[now] = max(back[now], temp);
		}
		for (int i = 0; i < g[now].size(); i++){
			q.push(g[now][i].to);
		}
	}
}

void start(){
	dfs();
	for (int i = 1; i <= n; i++){
		h[i] = 0;
	}

	vector<Node>gg;
	int now, to;
	for (now = 1; now <= n; now++){
		gg.clear();
		for (int i = 0; i < g[now].size(); i++){
			to = g[now][i].to;
			if (to != t[now]){
				gg.push_back(g[now][i]);
			}
		}
		g[now] = gg;
	}

	for (int x = 1; x < df; x++){
		now = dfv[x];
		for (int i = 0; i < g[now].size(); i++){
			to = g[now][i].to;
			h[now] = max(h[now], h[to] + 1);
		}
	}

	for (now = 1; now <= n; now++){
		for (int i = 0; i < g[now].size(); i++){
			to = g[now][i].to;
			g[now][i].h = h[to] + 1;
		}
	}

	for (now = 1; now <= n; now++){
		sort(g[now].begin(), g[now].end());
	}

	int id;
	int len = -1;
	int temp;
	for (now = 1; now <= n; now++){
		temp = find_len(now);
		if (temp > len){
			len = temp;
			id = now;
		}
	}

	for (int i = 1; i <= n; i++){
		isline[i] = false;
	}
	line.clear();
	get_line(id);
	
	find_back();

	for (int i = 0; i < line.size(); i++){
		a[i] = max_way(line[i]);
		if (line[i] == id){
			a[i] = max(a[i], back[id]);
		}
	}
	
	sparse_table();

	int ans=inf;
	int left;
	int right;
	int l, r;
	for (int lend = 0; lend < line.size(); lend++){
		l = lend;
		r = line.size() - 1;
		while (l + 2 < r){
			int mid = (l + r) / 2;
			if (find(lend, mid) > (int)line.size() - 1 - mid){
				r = mid;
			}
			else{
				l = mid;
			}

		}
		for (int x = l; x <= r; x++){
			temp = max(find(lend, x), (int)line.size() - 1 - x);
			temp = max(temp, lend);
			if (temp < ans){
				ans = temp;
				left = lend;
				right = x;
			}
		}
	}
	cout << ans << " " << line[left] << " " << line[right] << endl;
}

int main(){
	for (int i = 0; i < 20; i++){
		if ((1 << i) < maxn){
			lg2[1 << i] = i;
		}
	}
	for (int i = 3; i < maxn; i++){			
		if (!lg2[i]){
			lg2[i] = lg2[i - 1];
		}
	}

	int T;
	cin >> T;
	while (T--){
		cin >> n;
		for (int i = 1; i <= n; i++){
			g[i].clear();
		}
		node.h = 0;
		for (int i = 1; i <= n - 1; i++){
		//	cin >> node.now >> node.to;
			SS(node.now); SS(node.to);
			g[node.now].push_back(node);
			swap(node.now, node.to);
			g[node.now].push_back(node);
		}
		start();
	}
	return 0;
}

ZOJ 3820 2014ACM/ICPC牡丹江赛区B题

标签：style   blog   io   os   ar   for   sp   2014   art   

原文地址：http://blog.csdn.net/zz_1215/article/details/40024197