# Google Kickstart 2021 Round A My Editorial Published

on | Chinese version

First, you need to calculate the current score \$\$ K_c \$\$. The answer is \$\$ \$ | K_c-K | \$\$ because you can increase or decrease your score by \$\$ \$ 1 \$\$ in a single operation.

The time complexity is \$\$ \$ mathcal {O} (| S |) \$\$. The complexity of the space is \$\$ \$ mathcal {O} (1) \$\$. Code (C ++) # include
#include

Use namespace std.template void read (T & x) {x = 0; char c = getchar (); T sig = 1; for (;! Isdigit (c); c = getchar ()) if (c ==’-‘) sig = -1 ; for (; isdigit (c); c = getchar ()) x = (x << 3) + (x< 1) + c — '0'; x *= sig; }class Solution { public: void solve(int case_num) { printf("Case #%d: ", case_num); int n, k; read(n), read(k); string s; cin >> s; int cnt = 0; for (int i = 0; i[i] !! = s[n — 1 — i]) Cnt ++;} printf (“% d n”, abs (cnt — k));}}; int main () {int t; read

First, we use a simple dynamic programming method, namely \$\$ W, to calculate the length of the longest continuous \$\$ 1 \$\$ starting from each cell.[i][j]\$\$, \$\$ E[i][j]\$\$, \$\$ N[i][j]\$\$ and \$\$ S[i][j]\$\$.

Then enumerate each cell and count the L-shaped plot centered on the cell. There are four types of directional L-shaped plots.

And for each type, you need to consider which side is the longer side.

The time complexity is \$\$ \$ mathcal {O} (RC) \$\$. The complexity of the space is \$\$ \$ mathcal {O} (RC) \$\$. Code (C ++) # include
#include
#include

Use namespace std. Use ll = long long.template void read (T & x) {x = 0; char c = getchar (); T sig = 1; for (;! Isdigit (c); c = getchar ()) if (c ==’-‘) sig = -1 ; for (; isdigit (c); c = getchar ()) x = (x << 3) + (x << 1) + c — '0'; x * = sig;} class Solution {public: void resolve (Int case_num) {printf ("Case #% d:", case_num); int r, c; read (r), read (c); vector> m (r, vector(C)); for (int i = 0; i[i][j]); Vector> W (r, vector(C)); Vector> E (r, vector(C)); Vector> N (r, vector(C)); Vector> S (r, vector(C)); for (int i = 0; i < c; ++j) { if (m[i][j]) { W[i][j] = 1; if (j >= 1) W[i][j] + = W[i][j — 1];}} for (int j = c — 1; j> = 0; –j) {if (m)[i][j]) {E[i][j] = 1; if (j + 1[i][j] + = E[i][j + 1];}}} for (int j = 0; j < r; ++i) { if (m[i][j]) { N[i][j] = 1; if (i >= 1) N[i][j] + = N[i — 1][j];}} for (int i = r — 1; i> = 0; –i) {if (m)[i][j]) {S[i][j] = 1; if (i + 1[i][j] + = S[i + 1][j];}}} ll ans = 0; for (int i = 0; i[i][j], N[i][j] / 2) — 1); ans + = max (0, min (W)[i][j] / 2, N[i][j]) — 1); ans + = max (0, min (W)[i][j], S[i][j] / 2) — 1); ans + = max (0, min (W)[i][j] / 2, S[i][j]) -1); Year + = Maximum (0, Minimum (E)[i][j], N[i][j] / 2) -1); Year + = Maximum (0, Minimum (E)[i][j] / 2, N[i][j]) -1); Year + = Maximum (0, Minimum (E)[i][j], S[i][j] / 2) -1); Year + = Maximum (0, Minimum (E)[i][j] / 2, S[i][j]) — 1);} printf (“% lld n”, ans);}}; int main () {ios :: sync_with_stdio (false); cin.tie (0); int t; read

Obviously, you shouldn’t increase the highest cells. This insight can solve this problem in a Dijkstra-like way.

Use max-heap to store the cells to process. When processing cells with a height of \$\$ h \$\$, you need to make sure that all adjacent cells are at least as high as \$\$ h-1 \$\$. If the height of an adjacent cell increases, that cell must be queued at the new height.

Finally, get the answer by accumulating the difference between the final height and the original height of each cell.

The time complexity is \$\$ \$ mathcal {O} (RC log (RC)) \$\$. The complexity of the space is \$\$ \$ mathcal {O} (RC log (RC)) \$\$. Code (C ++) # include
#include
#include
#include
#include

Use namespace std. Use ll = long long.template void read (T & x) {x = 0; char c = getchar (); T sig = 1; for (;! Isdigit (c); c = getchar ()) if (c ==’-‘) sig = -1 ; for (; isdigit (c); c = getchar ()) x = (x << 3) + (x << 1) + c — '0'; x * = sig;} const int d = {{-1, 0}, {0, -1}, {1, 0}, {0, 1}}; class Solution {public: void resolve (int case_num) {printf ("Case #% d:" , Case_num); int r, c; read (r), read (c); vector> g (r, vector(C)); priority_queue> pq; for (int i = 0; i[i][j]); pq.emplace (g[i][j], I, j);} vector> g0 (g); while (! Pq.empty ()) {auto [h, i, j] = pq.top (); pq.pop (); if (h = r || nj < 0 || nj >= c) Continue. if (h — 1> g[ni][nj]) {Pq.emplace (h-1, ni, nj); g[ni][nj] = h — 1;}}} ll ans = 0; for (int i = 0; i[i][j] — G0[i][j]printf (“% lld n”, ans);}}; int main () {ios :: sync_with_stdio (false); cin.tie (0); int t; read

The hardest part of this problem is recognizing that it is actually a problem with the smallest spanning tree (actually the largest spanning tree, but equivalent).

The first observation is that at the cost of \$\$ 0 \$\$, we consider the known number to be unknown and can solve the equivalent problem where all the numbers are unknown.

The second observation is that the \$\$ N times N \$\$ matrix \$\$ (N-1) ^ 2 \$\$ cell needs to be checked accurately.

If you check less than \$\$ (N-1) ^ 2 \$\$, at least the cells of \$\$ 2N \$\$ are unknown, but up to the cells of \$\$ 2N-1 \$\$ You can guide. \$\$ (N-1) ^ 2 When checking more than \$\$, there are at least some rows / columns where all cells are checked, but this is not necessary.

Now, instead of determining the \$\$ (N-1) ^ 2 \$\$ cell to check, we try to determine the \$\$ 2N-1 \$\$ cell to be guided. You need to maximize the total cost so that you can minimize the total of the checked cells.

The insight is when you consider the \$\$ N \$\$ rows and the \$\$ N \$\$ columns as nodes and the cells as undirected edges (cells \$\$ (i, j) \$\$). Is considered to be an edge \$\$ r_i leftrightarrow c_j \$\$), and while you have to select the \$\$ 2N-1 \$\$ edge from all edges, exactly the \$\$ 2N \$\$ node Exists. So that the sum of those weights is maximized. Very close to MST!

To make sure it’s exactly MST, you need to check if loops are allowed. The loop must be in the form \$\$ \$ r_0 leftrightarrow c_0 leftrightarrow r_1 leftrightarrow c_1 cdots leftrightarrow r_0 \$\$ because there are no edges connecting two rows or two columns. If a loop occurs, every row and column in this loop will have at least two unknown cells, which cannot be resolved.

Therefore, you must select the \$\$ 2N-1 \$\$ Edge in the \$\$ 2N \$\$ node and graphs where loops are not allowed. This has been confirmed to be MST.

The MST part is very simple and both Kruskal and Prim work fluently.

The time complexity is \$\$ \$ mathcal {O} (N ^ 2 log N) \$\$. The complexity of the space is \$\$ \$ mathcal {O} (N ^ 2 log N) \$\$. Code (C ++) # include
#include
#include
#include

Use namespace std. Use ll = long long.template void read (T & x) {x = 0; char c = getchar (); T sig = 1; for (;! Isdigit (c); c = getchar ()) if (c ==’-‘) sig = -1 ; for (; isdigit (c); c = getchar ()) x = (x << 3) + (x << 1) + c — '0'; x * = sig;} Class UnionFind {int n; Vector Parent, size; public: UnionFind (int n) {this-> n = n; parent = vector(N); size = vector(N, 1); for (int i = 0; i < n; ++i) parent[i] = i; }int find(int idx) { if (parent[idx] == idx) return idx; return parent[idx] = find(parent[idx]); }void connect(int a, int b) { int fa = find(a), fb = find(b); if (fa != fb) { if (size[fa] > size[fb]){parent[fb] = fa; size[fa] + = Size[fb];} else {parent[fa] = fb; size[fb] + = Size[fa];}}}}; class Solution {public: void resolve (int case_num) {printf (“Case #% d:”, case_num); int n; read (n); Vector> a (n, vector(N)); Vector> b (n, vector(N)); Vector r (n); vector c (n); for (int i = 0; i[i][j]); Vector> Edge; ll tot = 0; for (int i = 0; i[i][j]); tot + = b[i][j]Edges.emplace_back (b[i][j], I, n + j);} sort (edges.rbegin (), edges.rend ()); UnionFind uf (n * 2); for (int i = 0; i[i]); For (int i = 0; i[i]); ll remove = 0; for (auto) [weight, u, v] : Edge) {if (uf.find (u) == uf.find (v)) continue; + = Remove weight. uf.connect (u, v);} printf (“% lld n”, tot — delete);}}; int main () {ios :: sync_with_stdio (false); cin.tie (0); int t; read

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