# TopCoder SRM 704 Div1 Easy: TreeDistanceConstruction

,

AGC 005 Cがこれの存在判定だけするものだったらしい。

## problem

• 任意の頂点$i$について、その頂点から最も遠い頂点を$j$としたとき、距離$d(i,j) = d_i$

## solution

Fix the centeral vertex(s) and put others to it in the ascending order of $d_i$. $O(N)$.

Let $C = \operatorname{argmin}_i d_i$, then $|C| \le 2$ holds or to construct is impossible. Make the centeral vertex/edge from $C$, and put other vertices to the both side of it in the ascending order of $d_i$. Validating the constructed tree with $O(N^3)$ helps you to avoid corner cases.

## implementation

#include <bits/stdc++.h>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
#define repeat_from(i,m,n) for (int i = (m); (i) < (n); ++(i))
#define whole(f,x,...) ([&](decltype((x)) whole) { return (f)(begin(whole), end(whole), ## __VA_ARGS__); })(x)
using namespace std;
template <class T> void setmax(T & a, T const & b) { if (a < b) a = b; }
template <class T> void setmin(T & a, T const & b) { if (b < a) a = b; }
const int inf = 1e9+7;
class TreeDistanceConstruction { public: vector<int> construct(vector<int> d); };

vector<int> TreeDistanceConstruction::construct(vector<int> d) {
int n = d.size();
vector<int> result;
int min_d = *whole(min_element, d);
int cnt_min_d = whole(count, d, min_d);
if (cnt_min_d == 1 or cnt_min_d == 2) { // construct
int l = whole(find, d, min_d) - d.begin();
int r;
if (cnt_min_d == 1) {
r = l;
} else {
r = find(d.begin() + l + 1, d.end(), min_d) - d.begin();
result.push_back(l);
result.push_back(r);
}
int m = l;
int tick = 0;
vector<int> que(n);
whole(iota, que, 0);
whole(sort, que, [&](int i, int j) { return d[i] < d[j]; });
repeat_from (top, cnt_min_d, n) {
int i = que[top];
if (d[m] + 1 < d[i]) {
tick = 0;
m = l;
}
if (tick == 0) {
result.push_back(l);
result.push_back(i);
l = i;
} else if (tick == 1) {
result.push_back(r);
result.push_back(i);
r = i;
} else {
result.push_back(m);
result.push_back(i);
}
++ tick;
}
} else {
// impossible
}
if (not result.empty()) { // validate
vector<vector<int> > dist(n, vector<int>(n, inf));
repeat (i,n) dist[i][i] = 0;
repeat (i,n-1) {
int u = result[2*i  ];
int v = result[2*i+1];
dist[u][v] = 1;
dist[v][u] = 1;
}
repeat (k,n) { // warshall floyd
repeat (i,n) {
repeat (j,n) {
setmin(dist[i][j], dist[i][k] + dist[k][j]);
}
}
}
vector<int> eccentricity(n);
repeat (i,n) {
repeat (j,n) {
setmax(eccentricity[i], dist[i][j]);
}
}
repeat (i,n) {
if (d[i] != eccentricity[i]) {
result.clear();
}
}
}
return result;
}