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8/11/2018 - 10:09 AM

## 103

Longest increasing subsequence problem, with a twist: elements are arrays of integers

``````// Problem #    : 103
// Created on   : 2018-08-11 02:23:17

#include <bits/stdc++.h>

#define REP(i, n) for (int i = 0; i < (n); ++i)
#define FOR(i, c)                                                              \
for (__typeof((c).begin()) i = (c).begin(); i != (c).end(); ++i)
#define ALL(c) (c).begin(), (c).end()
#define UNIQUE(c) (c).resize(unique(ALL(c)) - (c).begin())
#define SZ(x) ((int)((x).size()))

using namespace std;

typedef long long ll;
typedef pair<int, int> ii; // pair of ints
typedef vector<int> vi;    // 1d vector of ints
typedef vector<ii> vii;    // 1d vector of pairs
typedef vector<vi> vvi;    // 2d vector of ints
typedef vector<vii> vvii;  // 2d vector of pairs

int V, k;
vvi g;

// store parent information for each node, allows us to reconstruct the elements
// that compose the LIS
vi p;

// Returns false if ANY element of g[x][i], for i in 0..k-1, is less than or
// equal to the corresponding element of g[y][i].  In other words, true if 1st
// argument is strictly GREATER than 2nd argument
// (i.e. 'y' can 'fit' inside 'x')
bool cmp(int x, int y) {
for (int i = 0; i < k; i++) {
if (g[x][i] <= g[y][i])
return false;
}
return true;
}

void path(int i) {
if (i == p[i]) {
cout << g[i].back() + 1;
return;
}
path(p[i]);
cout << " " << g[i].back() + 1;
}

void LIS() {
vi b(V, 1);
p.assign(V, 0);
iota(ALL(p), 0);

for (int i = 1; i < V; i++) {
for (int j = 0; j < i; j++) {
if (cmp(i, j) && b[i] < b[j] + 1) {
b[i] = b[j] + 1, p[i] = j;
}
}
}

int mx = 0, idx;
REP(i, V) {
if (b[i] > mx) {
mx = b[i], idx = i;
}
}

cout << mx << "\n";
path(idx);
cout << "\n";
}

int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);

while (cin >> V >> k) {
g.assign(V, vi(k));
REP(i, V) {
REP(j, k) { cin >> g[i][j]; }
sort(g[i].begin(), g[i].end());

// push the identifier onto the back of this array "item", to allow us to
// trace back through parents and get the correct representative item for
// output.
g[i].push_back(i);
}

sort(ALL(g), [](const vi &x, const vi &y) {
return lexicographical_compare(ALL(x), ALL(y));
});

LIS();
}

return 0;
}
``````