POJ 2296 Map Labeler(2-sat)
POJ 2296 Map Labeler
题意:
坐标轴上有N个点。要在每一个点上贴一个正方形,这个正方形的横竖边分别和x,y轴平行,而且要使得点要么在正方形的上面那条边的中点,或者在以下那条边的中点。而且随意两个点的正方形都不重叠(能够重边)。问正方形最大边长能够多少?
思路:显然的2-sat问题,注意推断两个矩形相交的地方,细节
代码:
#include <cstdio> #include <cstring> #include <cstdlib> #include <vector> #include <algorithm> using namespace std; const int MAXNODE = 205; struct TwoSet { int n; vector<int> g[MAXNODE * 2]; bool mark[MAXNODE * 2]; int S[MAXNODE * 2], sn; void init(int tot) { n = tot * 2; for (int i = 0; i < n; i += 2) { g[i].clear(); g[i^1].clear(); } memset(mark, false, sizeof(mark)); } void add_Edge(int u, int uval, int v, int vval) { u = u * 2 + uval; v = v * 2 + vval; g[u^1].push_back(v); g[v^1].push_back(u); } void delete_Edge(int u, int uval, int v, int vval) { u = u * 2 + uval; v = v * 2 + vval; g[u^1].pop_back(); g[v^1].pop_back(); } bool dfs(int u) { if (mark[u^1]) return false; if (mark[u]) return true; mark[u] = true; S[sn++] = u; for (int i = 0; i < g[u].size(); i++) { int v = g[u][i]; if (!dfs(v)) return false; } return true; } bool solve() { for (int i = 0; i < n; i += 2) { if (!mark[i] && !mark[i + 1]) { sn = 0; if (!dfs(i)){ for (int j = 0; j < sn; j++) mark[S[j]] = false; sn = 0; if (!dfs(i + 1)) return false; } } } return true; } } gao; const int N = 105; int t, n; struct Point { int x, y; void read() { scanf("%d%d", &x, &y); x *= 2; } } p[N]; bool judge(int len) { gao.init(n); for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { if (p[i].x + len <= p[j].x - len || p[j].x + len <= p[i].x - len) continue; for (int x = 0; x < 2; x++) { for (int y = 0; y < 2; y++) { int y1, y2, y3, y4; if (x == 0) { y1 = p[i].y - len; y2 = p[i].y; } else { y1 = p[i].y; y2 = p[i].y + len; } if (y == 0) { y3 = p[j].y - len; y4 = p[j].y; } else { y3 = p[j].y; y4 = p[j].y + len; } if ((y1 >= y3 && y1 < y4) || (y2 > y3 && y2 <= y4) || (y3 >= y1 && y3 < y2) || (y3 > y2 && y4 <= y2)) gao.add_Edge(i, x, j, y); } } } } return gao.solve(); } int main() { scanf("%d", &t); while (t--) { scanf("%d", &n); for (int i = 0; i < n; i++) p[i].read(); int l = 0, r = 20000; while (l < r) { int mid = (l + r) / 2; if (judge(mid)) l = mid + 1; else r = mid; } printf("%d\n", l - 1); } return 0; }