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UVA 11930 - Rectangles(2-sat + 计算几何)
UVA 11930 - Rectangles
题目链接
题意:给定一些矩形,每个在两条对角线选一条,保证全部不相交,问可不可行(这题有坑啊,矩形不一定平行坐标轴。。。)
思路:2-sat,主对角线为true,副对角线为false,枚举两个矩形的每条对角线,利用叉积判相交,如果相交就加一条边进去,最后2-sat判定即可
代码:
#include <cstdio> #include <cstring> #include <cstdlib> #include <vector> #include <algorithm> using namespace std; const int MAXNODE = 1005; 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; typedef long long ll; struct Point { ll x, y; Point() {} Point(ll x, ll y) { this->x = x; this->y = y; } void read() { scanf("%lld%lld", &x, &y); } }; int xmul(Point a, Point b, Point c) { ll ans = (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x); if (ans == 0) return 0; if (ans > 0) return 1; return -1; } bool judge(Point a1, Point b1, Point a2, Point b2) { if (min(a1.x, b1.x) <= max(a2.x, b2.x) && min(a1.y, b1.y) <= max(a2.y, b2.y) && min(a2.x, b2.x) <= max(a1.x, b1.x) && min(a2.y, b2.y) <= max(a1.y, b1.y) && xmul(a1, a2, b1) * xmul(a1, b2, b1) <= 0 && xmul(a2, a1, b2) * xmul(a2, b1, b2) <= 0) return true; else return false; } const int N = 1005; const ll INF = 0x3f3f3f3f; bool cmp(Point a, Point b) { if (a.x == b.x) return a.y < b.y; return a.x < b.x; } int n; struct Rec { Point a[2], b[2]; Rec() {} Rec(ll l, ll r, ll u, ll d) { a[0] = Point(l, u); b[0] = Point(r, d); a[1] = Point(r, u); b[1] = Point(l, d); } void init() { Point tmp[4]; for (int i = 0; i < 4; i++) tmp[i].read(); sort(tmp, tmp + 4, cmp); b[1] = tmp[0]; a[0] = tmp[1]; b[0] = tmp[2]; a[1] = tmp[3]; /* if (tmp[0].x != tmp[1].x) while(1); if (tmp[2].x != tmp[3].x) while(1); if (tmp[0].y != tmp[2].y) while(1); if (tmp[1].y != tmp[3].y) while(1); */ } } rec[N]; int main() { while (~scanf("%d", &n) && n) { gao.init(n); //ll x, y; //ll l, r, u, d; for (int i = 0; i < n; i++) { //l = d = INF; r = u = -INF; /*for (int j = 0; j < 4; j++) { scanf("%lld%lld", &x, &y); l = min(l, x); r = max(r, x); d = min(d, y); u = max(u, y); } rec[i] = Rec(l, r, u, d);*/ rec[i].init(); for (int j = 0; j < i; j++) { for (int x = 0; x < 2; x++) for (int y = 0; y < 2; y++) { if (judge(rec[i].a[x], rec[i].b[x], rec[j].a[y], rec[j].b[y])) gao.add_Edge(i, x, j, y); } } } printf("%s\n", gao.solve() ? "YES" : "NO"); } return 0; }
UVA 11930 - Rectangles(2-sat + 计算几何)
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