UVA 1146 - Now or later
题目链接
题意：n个飞机，每个飞机有一个早到时间和一个晚到时间，问怎么安排飞机，使得飞机到的间隔的最小值最大
思路：二分答案，然后利用2-set去判断，如果两个飞机的两个时刻间隔比这个时刻小，那么就是表示不能同时满足这两个条件，就加一条xi^xj的边进去，然后利用2-SET判定一下
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;
const int N = 2005;
int n, ti[N][2];
int L, R;
bool judge(int len) {
gao.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(ti[i][a] - ti[j][b]) < len)
gao.add_Edge(i, a^1, j, b^1);
}
}
}
}
return gao.solve();
}
int main() {
while (~scanf("%d", &n)) {
L = R = 0;
for (int i = 0; i < n; i++)
for (int j = 0; j < 2; j++) {
scanf("%d", &ti[i][j]);
R = max(R, ti[i][j]);
}
R++;
while (L < R) {
int mid = (L + R) / 2;
if (judge(mid)) L = mid + 1;
else R = mid;
}
printf("%d
", L - 1);
}
return 0;
}