POJ 3905 Perfect Election(2-sat)
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POJ 3905 Perfect Election
思路:非常裸的2-sat,就依据题意建边就可以
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); } 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; int n, m; int main() { while (~scanf("%d%d", &n, &m)) { gao.init(n); int u, v; while (m--) { scanf("%d%d", &u, &v); if (u > 0 && v > 0) { u--; v--; gao.add_Edge(u, 1, v, 1); } else if (u > 0 && v < 0) { v = -v; u--; v--; gao.add_Edge(u, 1, v, 0); } else if (u < 0 && v > 0) { u = -u; u--; v--; gao.add_Edge(u, 0, v, 1); } else if (u < 0 && v < 0) { u = -u; v = -v; u--; v--; gao.add_Edge(u, 0, v, 0); } } printf("%d\n", gao.solve()); } return 0; }
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