CF-85E.Guard Towers(二分+染色)

題目鏈接

題意

$n$個燈塔分成2分,求出最小的曼哈頓距離和其方案樹

題意

二分+染色

二分枚舉最小值$mid$,判斷能否將$n$個燈塔分為最大曼哈頓值不超過$mid$的兩份

方案數=$2^{聯通塊個數}$

#include <bits/stdc++.h> const int maxn = 5e3 + 5; const int mod = 1e9 + 7; using namespace std; pair<int,int> a[maxn]; int cnt, vis[maxn], n; long long Pow(long long a, long long b) {long long ans = 1;while (b) {if (b & 1) {ans *= a;ans %= mod;}a *= a;a %= mod;b >>= 1;}return ans; } int dfs(int u, int c, int mid) {for (int i = 0; i < n; ++i) {int len = abs(a[i].first - a[u].first) + abs(a[i].second - a[u].second);if (len <= mid || vis[i] == (c^1)) continue;if (vis[i] == c) return -1;vis[i] = c ^ 1;if (dfs(i, c^1, mid) == -1) return -1; }return 1; } int check(int mid) {fill(vis, vis+maxn, -1);cnt = 0;for (int i = 0; i < n; ++i) {if (vis[i] == -1) {cnt++;vis[i] = 0;if (dfs(i, 0, mid) == -1) return -1;} } } int main() {scanf("%d", &n);for (int i = 0; i < n; ++i) {scanf("%d %d", &a[i].first, &a[i].second);}int l = 0, r = 1e4;while (l <= r) {int mid = (l + r) >> 1;if (check(mid) == -1) l = mid + 1;else r = mid - 1; }check(l);printf("%d\n%lld\n", l, Pow(2, cnt));return 0; }

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