标签:style http color os io ar for 代码 sp
题意:一个城市,有南北和东西街道,每种街道都是单行道,现在给定几个起点和终点,要求起点和终点必须最多转一次弯可以到达,问是否可以满足所有的起点终点
思路:2-set,对于东西,南北街道,向西(北)为false,向东(南)为true,对于每个起点终点,肯定是走坐标对应的那4条街道,表示出来是(s1 & a2) | (s2 & a1),可以转化成(s1 | s2) & (s1 | a1) & (a2 | v2) & (a2 | a1)对应2-set4条边,注意如果横坐标或纵坐标相同,只要建一条边(s, s)即可,这样建图,进行2-set判定即可
代码:
#include <cstdio> #include <cstring> #include <cstdlib> #include <vector> #include <algorithm> using namespace std; const int MAXNODE = 2005; 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); } 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; int n, S, A, m; int main() { scanf("%d", &n); while (n--) { scanf("%d%d%d", &S, &A, &m); gao.init(S + A); int s1, a1, s2, a2; while (m--) { scanf("%d%d%d%d", &s1, &a1, &s2, &a2); s1--; a1--; s2--; a2--; if (s1 == s2 && a1 == a2) continue; a1 += S; a2 += S; if (s1 == s2) gao.add_Edge(s1, (a1 < a2), s2, (a1 < a2)); else if (a1 == a2) gao.add_Edge(a1, (s1 < s2), a2, (s1 < s2)); else { gao.add_Edge(s1, (a1 < a2), s2, (a1 < a2)); gao.add_Edge(s1, (a1 < a2), a1, (s1 < s2)); gao.add_Edge(a2, (s1 < s2), s2, (a1 < a2)); gao.add_Edge(a2, (s1 < s2), a1, (s1 < s2)); } } printf("%s\n", gao.solve() ? "Yes" : "No"); } return 0; }
标签:style http color os io ar for 代码 sp
原文地址:http://blog.csdn.net/accelerator_/article/details/38960541