LA3211 Now or later

题目大意：n架飞机，每架可选择两个着落时间。安排一个着陆时间表，使得着陆间隔的最小值最大。（转自http://blog.csdn.net/u013514182/article/details/42333363）

每个飞机有两个选择：时间1或时间2，分别用xi和x'表示。最小值最大，考虑二分答案ans;

对于任意两家飞机x,y,x选择时间k，y选择时间l，如果时间差小于ans，说明1、x选k后y必须不能选l，2、y选l后x必须不能选k

边开小了，疯狂RE，边居然要开到千万级别。。。竟然出了这么极端的数据专门卡空间。。。出题人真XX

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <queue>
 #include <vector>
 #include <map>
 #include <string>
 #include <cmath>
 #define min(a, b) ((a) < (b) ? (a) : (b))
 #define max(a, b) ((a) > (b) ? (a) : (b))
 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
 template<class T>
 inline void swap(T &a, T &b)
 {
     T tmp = a;a = b;b = tmp;
 }
 inline void read(int &x)
 {
     x = ;char ch = getchar(), c = ch;
     while(ch < '' || ch > '') c = ch, ch = getchar();
     while(ch <= '' && ch >= '') x = x *  + ch - '', ch = getchar();
     if(c == '-') x = -x;
 }
 const int INF = 0x3f3f3f3f;
 const int MAXN =  + ;
 struct Edge
 {
     int u,v,nxt;
     Edge(int _u, int _v, int _nxt){u = _u;v = _v;nxt = _nxt;}
     Edge(){}
 }edge[];
 int head[MAXN], cnt;
 inline void insert(int a, int b)
 {
     edge[++ cnt] = Edge(a, b, head[a]), head[a] = cnt;
 }
 int n, t[MAXN][], dfn[MAXN], dfst, low[MAXN], b[MAXN], bb[MAXN], group, belong[MAXN], stack[MAXN], top;
 void dfs(int u)
 {
     b[u] = bb[u] = , stack[++ top] = u, dfn[u] = low[u] = ++ dfst;
     for(int pos = head[u];pos;pos = edge[pos].nxt)
     {
         int v = edge[pos].v;
         if(!b[v]) dfs(v), low[u] = min(low[v], low[u]);
         else if(bb[v]) low[u] = min(low[u], dfn[v]);
     }
     if(low[u] == dfn[u])
     {
         ++ group;
         int now = -;
         while(now != u)    now = stack[top --], belong[now] = group, bb[now] = ;
     }
 }

 void tarjan()
 {
     dfst = , group = , memset(belong, , sizeof(belong)), memset(dfn, , sizeof(dfn)), memset(low, , sizeof(low)), memset(b, , sizeof(b)), memset(bb, , sizeof(bb));
     for(int i = ;i <= n;++ i) if(!b[i << ]) dfs(i << );
     for(int i = ;i <= n;++ i) if(!b[i <<  | ]) dfs(i <<  | );
 }

 int check(int m)
 {
     memset(head, , sizeof(head)), cnt = ;
     for(int i = ;i <= n;++ i)
         for(int k = ;k <= ;++ k)
             for(int j = i + ;j <= n;++ j)
                 for(int l = ; l <= ;++ l)
                     if(abs(t[i][k] - t[j][l]) < m)
                         insert(i <<  | k, j <<  | (l ^ )), insert(j <<  | l, i <<  | (k ^ ));
     tarjan();
     for(int i = ;i <= n;++ i) if(belong[i << ] == belong[i <<  | ]) return ;
     return ;
 }

 int main()
 {
     while(scanf("%d", &n) != EOF)
     {
         int l = , r = , mid, ans = ;
         for(int i = ;i <= n;++ i) read(t[i][]), read(t[i][]), r = max(r, max(t[i][], t[i][]));
         while(l <= r)
         {
             mid = (l + r) >> ;
             if(check(mid)) l = mid + , ans = mid;
             else r = mid - ;
         }
         printf("%d\n", ans);
     }
     return ;
 }

LA3211