ZOJ 3820 2014ACM/ICPC牡丹江司B称号

3797714 2014 - 10 - 12 21:58 : 19 Accepted 3820 C++ 1350 70240 zz_1215
比較麻烦的一道题吧,開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,能够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一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,吐槽下这样的考代码能力的题

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。由于这个我又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
"
#define in64 "%I64d"
#else
#define i64 long long
#define out64 "%lld
"
#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; }



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原文地址:https://www.cnblogs.com/mfrbuaa/p/4738716.html