标签:style blog http color io os ar for sp
题意:N串钥匙,每串2把,只能选一把,然后有n个大门,每个门有两个锁,开了一个就能通过,问选一些钥匙,最多能通过多少个门
思路:二分通过个数,然后对于钥匙建边至少一个不选,门建边至少一个选,然后2-sat搞一下即可。
一开始是按每串钥匙为1个结点,可是后面发现数据有可能一把钥匙,出现在不同串(真是不合理),所以这个做法就跪了
代码:
#include <cstdio> #include <cstring> #include <cstdlib> #include <vector> #include <algorithm> using namespace std; const int MAXNODE = 2105; 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 = 2055; int n, m; int x[N], y[N]; int k1[N], k2[N]; bool judge(int d) { gao.init(2 * n); for (int i = 0; i < n; i++) gao.add_Edge(x[i], 0, y[i], 0); for (int i = 1; i <= d; i++) gao.add_Edge(k1[i], 1, k2[i], 1); return gao.solve(); } int main() { while (~scanf("%d%d", &n, &m) && n) { for (int i = 0; i < n; i++) scanf("%d%d", &x[i], &y[i]); for (int i = 1; i <= m; i++) scanf("%d%d", &k1[i], &k2[i]); int l = 0, r = m + 1; while (l < r) { int mid = (l + r) / 2; if (judge(mid)) l = mid + 1; else r = mid; } printf("%d\n", l - 1); } return 0; }
HDU 1816, POJ 2723 Get Luffy Out(2-sat)
标签:style blog http color io os ar for sp
原文地址:http://blog.csdn.net/accelerator_/article/details/40476057