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
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <algorithm> 6 #include <queue> 7 #include <vector> 8 #include <map> 9 #include <string> 10 #include <cmath> 11 #define min(a, b) ((a) < (b) ? (a) : (b)) 12 #define max(a, b) ((a) > (b) ? (a) : (b)) 13 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a)) 14 template<class T> 15 inline void swap(T &a, T &b) 16 { 17 T tmp = a;a = b;b = tmp; 18 } 19 inline void read(int &x) 20 { 21 x = 0;char ch = getchar(), c = ch; 22 while(ch < '0' || ch > '9') c = ch, ch = getchar(); 23 while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar(); 24 if(c == '-') x = -x; 25 } 26 const int INF = 0x3f3f3f3f; 27 const int MAXN = 50000 + 10; 28 struct Edge 29 { 30 int u,v,nxt; 31 Edge(int _u, int _v, int _nxt){u = _u;v = _v;nxt = _nxt;} 32 Edge(){} 33 }edge[20000000]; 34 int head[MAXN], cnt; 35 inline void insert(int a, int b) 36 { 37 edge[++ cnt] = Edge(a, b, head[a]), head[a] = cnt; 38 } 39 int n, t[MAXN][2], dfn[MAXN], dfst, low[MAXN], b[MAXN], bb[MAXN], group, belong[MAXN], stack[MAXN], top; 40 void dfs(int u) 41 { 42 b[u] = bb[u] = 1, stack[++ top] = u, dfn[u] = low[u] = ++ dfst; 43 for(int pos = head[u];pos;pos = edge[pos].nxt) 44 { 45 int v = edge[pos].v; 46 if(!b[v]) dfs(v), low[u] = min(low[v], low[u]); 47 else if(bb[v]) low[u] = min(low[u], dfn[v]); 48 } 49 if(low[u] == dfn[u]) 50 { 51 ++ group; 52 int now = -1; 53 while(now != u) now = stack[top --], belong[now] = group, bb[now] = 0; 54 } 55 } 56 57 void tarjan() 58 { 59 dfst = 0, group = 0, memset(belong, 0, sizeof(belong)), memset(dfn, 0, sizeof(dfn)), memset(low, 0, sizeof(low)), memset(b, 0, sizeof(b)), memset(bb, 0, sizeof(bb)); 60 for(int i = 1;i <= n;++ i) if(!b[i << 1]) dfs(i << 1); 61 for(int i = 1;i <= n;++ i) if(!b[i << 1 | 1]) dfs(i << 1 | 1); 62 } 63 64 int check(int m) 65 { 66 memset(head, 0, sizeof(head)), cnt = 0; 67 for(int i = 1;i <= n;++ i) 68 for(int k = 0;k <= 1;++ k) 69 for(int j = i + 1;j <= n;++ j) 70 for(int l = 0; l <= 1;++ l) 71 if(abs(t[i][k] - t[j][l]) < m) 72 insert(i << 1 | k, j << 1 | (l ^ 1)), insert(j << 1 | l, i << 1 | (k ^ 1)); 73 tarjan(); 74 for(int i = 1;i <= n;++ i) if(belong[i << 1] == belong[i << 1 | 1]) return 0; 75 return 1; 76 } 77 78 int main() 79 { 80 while(scanf("%d", &n) != EOF) 81 { 82 int l = 1, r = 0, mid, ans = 0; 83 for(int i = 1;i <= n;++ i) read(t[i][0]), read(t[i][1]), r = max(r, max(t[i][1], t[i][0])); 84 while(l <= r) 85 { 86 mid = (l + r) >> 1; 87 if(check(mid)) l = mid + 1, ans = mid; 88 else r = mid - 1; 89 } 90 printf("%d ", ans); 91 } 92 return 0; 93 }