UV 10319, uva10319

UV 10319, uva10319
Ultraviolet A 10319-Manhattan

Question Link

A city has north and south streets and east and west streets. Each Street is a one-way street. Now, given a few starting points and endpoints, the start and end points must be bent at most once to arrive, can all start and end points be satisfied?

Idea: 2-set. For east-west, north-south streets, false for West (north), true for East (south), and for each start point, it must be the four sub-streets corresponding to the coordinates. It indicates (s1 & a2) | (s2 & a1), which can be converted to (s1 | s2) & (s1 | a1) & (a2 | v2) & (a2 | a1) corresponds to 2-set4 edges. If the horizontal or vertical coordinates are the same, you only need to create an edge (s, s, perform 2-set determination.

Code:

#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;}