APIO 11-color

Author: thecodingwizard

APIO/APIO 11-Color.cpp

@@ -0,0 +1,109 @@
+// USACO Guide has a slightly different solution
+// Note that the entire grid is dependent on the first row and the first column
+// The cell (i, j) is dependent on (1, 1), (1, j), and (i, 1)
+// Therefore, make a graph: 0...n-1 represents (i, 1). n...n+m-1 represents(1, j).
+// For every cell that's already colored in, draw an edge between the row and the column nodes it's dependent on
+// Mark the edge as "these two nodes have the same value" or "these two nodes have different values" depending on the value of the cell
+// Set (1, 1) to red and calculate the # of valid assignments
+// Set (1, 1) to blue and calculate the # of valid assignments
+// To calculate # of valid assignments, we basically DFS from every node (beginning from the nodes whose values we already know).
+// Set the values of all connected nodes.
+// For nodes whose values we don't know, just arbitrarily assign a value and multiply the # of valid assignments by 2.
+
+#include <bits/stdc++.h>
+
+using namespace std;
+
+using ll = long long;
+#define ii pair<int, int>
+#define f first
+#define s second
+#define pb push_back
+#define mp make_pair
+#define all(x) x.begin(), x.end()
+#define sz(x) (int)x.size()
+#define F0R(i, n) for (int i = 0; i < n; i++)
+#define FOR(i, a, b) for (int i = a; i < b; i++)
+#define inf 1000000010
+
+// [0...n-1] is rows
+// [n...n+m-1] is columns
+// n+m is 0
+// n+m+1 is 1
+// second: 1 if not equal, 0 if equal
+vector<ii> adj[300000];
+int color[300000];
+bool vis[300000];
+int topLeftC;
+
+ll ans = 1;
+
+void dfs(int u) {
+    if (vis[u]) return;
+    vis[u] = true;
+    if (color[u] == -1) {
+        //cerr << u << endl;
+        ans = (ans*2)%((int)1e9);
+        color[u] = 0;
+    }
+    for (ii v : adj[u]) {
+        int expected = color[u] ^ v.s ^ topLeftC;
+        if (color[v.f] == -1) color[v.f] = expected;
+        if (color[v.f] != expected) {
+            ans = 0;
+        }
+        dfs(v.f);
+    }
+}
+
+int main() {
+    cin.tie(0)->sync_with_stdio(0);
+
+    int n, m, k; cin >> n >> m >> k;
+    int topLeft = -1;
+    color[n+m] = 0;
+    color[n+m+1] = 1;
+    for (int i = 0; i < k; i++) {
+        int x, y, c; cin >> x >> y >> c;
+        if (x == y && x == 1) {
+            topLeft = c;
+        } else if (x == 1) {
+            adj[n+y-1].emplace_back(n+m+c, 0);
+            adj[n+m+c].emplace_back(n+y-1, 0);
+        } else if (y == 1) {
+            adj[x-1].emplace_back(n+m+c, 0);
+            adj[n+m+c].emplace_back(x-1, 0);
+        } else {
+            if (x%2==0 && y%2==0) {
+                c ^= 1;
+            }
+            int r = x-1, col = n+y-1;
+            adj[r].pb(mp(col, c));
+            adj[col].pb(mp(r, c));
+        }
+    }
+
+    ll totAns = 0;
+    for (topLeftC = 0; topLeftC <= 1; topLeftC++) {
+        if (topLeft != -1 && topLeft != topLeftC) continue;
+        ans = 1;
+        for (int i = 0; i < n+m; i++) {
+            color[i] = -1;
+        }
+        for (int i = 0; i < n+m+10; i++) {
+            vis[i] = false;
+        }
+        dfs(n+m);
+        dfs(n+m+1);
+        for (int i = 0; i < n+m; i++) {
+            if (i == 0 || i == n) continue; // these point to the upper left cell that's a special case
+            dfs(i);
+        }
+        totAns = (totAns + ans) % ((int)1e9);
+    }
+
+
+    cout << totAns << endl;
+
+    return 0;
+}