UVA1146 Now or later
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一看要求一个最小的最大间隔,最小最大这类字眼,果断的就是二分枚举答案,然后进行验证,验证的时候就有一个炫酷的解法了,就是2-SAT判断是否矛盾,对于一个2-SAT问题,我们可以这么做,先用tarjan,然后判断同一个状态拆成的两个节点是否在一个连通分量内,如果在,表示矛盾,如果不在,表示成立,求可行解的方法就是重构一个有向无环图,按照拓扑序反向输出可行解。
或者说我们可以假设某一个点为真,然后进行染色,如果在染色过程中出现了该点的对立也被染色了,那么就表明此点为真是错误的,那么我们再换成此点为假重新染色,如果无法成功染色就证明该问题无解,如果染上了,那么被着色节点就是可行解。
#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <cstdlib>using namespace std;const int maxn = 44444;struct TwoSAT{int n, S[maxn], c;vector<int> G[maxn];bool mark[maxn];bool dfs(int x){if (mark[x ^ 1]) return 0;if (mark[x]) return 1;mark[x] = 1;S[c++] = x;for (int i = 0; i < G[x].size(); i++)if (!dfs(G[x][i])) return 0;return 1;}void init(int n){this -> n = n;for (int i = 0; i <= n * 2; i++)G[i].clear();memset(mark, 0, sizeof(mark));}void add_clause(int x, int xval, int y, int yval){x = x * 2 + xval;y = y * 2 + yval;G[x ^ 1].push_back(y);G[y ^ 1].push_back(x);}bool solve(){for (int i = 0; i < n * 2; i += 2)if (!mark[i] && !mark[i + 1]){c = 0;if (!dfs(i)){while(c > 0) mark[S[--c]] = 0;if (!dfs(i + 1)) return 0;}}return 1;}};int n, T[maxn][2];TwoSAT solver;bool check(int t){solver.init(n);for (int i = 0; i < n; i++) for (int a = 0; a < 2; a++)for (int j = i + 1; j < n; j++) for (int b = 0; b < 2; b++)if (abs(T[i][a] - T[j][b]) < t) solver.add_clause(i, a ^ 1, j, b ^ 1);if (solver.solve()) return 1;else return 0;}int main(){while(~scanf("%d", &n) && n){int l = 0, r = 0;for (int i = 0; i < n; i++)for (int j = 0; j < 2; j++){scanf("%d", &T[i][j]);r = max(r, T[i][j]);}while (l < r){int m = l + (r - l + 1) / 2;if (check(m)) l = m;else r = m - 1;}printf("%d\n", l);}return 0;}上面为刘汝佳解法,下面为tarjan一般解法
参考资料:伍昱《由对称性解2-SAT问题》,赵爽《2-SAT 解法浅析》,刘汝佳《算法竞赛入门经典训练指南》,但是佳哥的解法和本菜所说的不一样。
#include <cstdio>#include <cstring>#include <algorithm>#include <cstdlib>#include <vector>#include <stack>using namespace std;const int N = 4444;int pre[N], sccno[N], lowlink[N], dfs_clock, scc_cnt, T[N][2], n;stack<int> S;vector<int> G[N];void add_edge(int x, int xval, int y, int yval){ x = x * 2 + xval; y = y * 2 + yval; G[x ^ 1].push_back(y); G[y ^ 1].push_back(x);}void dfs(int u){ pre[u] = lowlink[u] = ++dfs_clock; S.push(u); for (int i = 0; i < G[u].size(); i++){ int v = G[u][i]; if (!pre[v]){ dfs(v); lowlink[u] = min(lowlink[u], lowlink[v]); } else if (!sccno[v]) lowlink[u] = min(lowlink[u], pre[v]); } if (lowlink[u] == pre[u]){ scc_cnt += 1; for (;;){ int x = S.top(); S.pop(); sccno[x] = scc_cnt; if (x == u) break; } }}void tarjan(int n){ dfs_clock = scc_cnt = 0; memset(pre, 0, sizeof(pre)); memset(sccno, 0, sizeof(sccno)); memset(lowlink, 0, sizeof(lowlink)); for (int i = 0; i < n; i++) if (!pre[i]) dfs(i);}bool check(int t){ for (int i = 0; i <= 2 * n; i++) G[i].clear(); for (int i = 0; i < n; i++) for (int a = 0; a < 2; a++) for (int j = i + 1; j < n; j++) for (int b = 0; b < 2; b++) if (abs(T[i][a] - T[j][b]) < t) add_edge(i, a ^ 1, j, b ^ 1); tarjan(n * 2); //for (int i = 0; i < n * 2; i++) printf("%d ", sccno[i]); //printf("\n"); for (int i = 0; i < n * 2; i += 2) if (sccno[i] == sccno[i + 1]) return 0; return 1;}int main(){ while(~scanf("%d", &n) && n){ int l = 0, r = 0; for (int i = 0; i < n; i++) for (int j = 0; j < 2; j++){ scanf("%d", &T[i][j]); r = max(r, T[i][j]); } while (l < r){ int m = l + (r - l + 1) / 2; if (check(m)) l = m; else r = m - 1; } printf("%d\n", l); } return 0;}
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