uva 1146(2-SAT+二分判断)

题意：有n架飞机需要着陆，每架飞机都有两个时间可以选择，早和晚。告诉你每架飞机的这两个时间，然后我们定义相邻两架飞机着陆时间间隔的最小值为安全间隔。问你安全间隔最大能为多少。

思路：一个典型的2-sat+二分判断的问题。首先我们对安全间隔二分答案，然后再2-sat判断可行性。思路理清的话没什么难度。

代码如下：

/**************************************************
 * Author     : xiaohao Z
 * Blog     : http://www.cnblogs.com/shu-xiaohao/
 * Last modified : 2014-02-01 12:12
 * Filename     : uva_1146.cpp
 * Description     : 
 * ************************************************/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<unsigned int,unsigned int> puu;
typedef pair<int, double> pid;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;

const int INF = 0x3f3f3f3f;
const double eps = 1E-6;
const int LEN = 10020;
vector<int> Map[LEN], rMap[LEN], vs;
int n, vis[LEN], tim[LEN][2], sccn[LEN];

int abs(int a){
     return a>0?a:-a;
}

void dfs(int v){
    vis[v] = 1;
    for(int i=0; i<Map[v].size(); i++)
       if(!vis[Map[v][i]])dfs(Map[v][i]);
    vs.PB(v);
}
 
void rdfs(int v, int f){
    vis[v] = 1;
    sccn[v] = f;
    for(int i=0; i<rMap[v].size(); i++)
        if(!vis[rMap[v][i]]) rdfs(rMap[v][i], f);
}

int scc()
{
     memset(vis, 0, sizeof vis);
     vs.clear();
     for(int i=0; i<n*2; i++) if(!vis[i])dfs(i);
     memset(vis, 0, sizeof vis);
     int k = 0;
     for(int i=vs.size()-1; i>=0; i--)if(!vis[vs[i]])rdfs(vs[i], k++);
     return k;
}

void addedge(int a, int b){
    Map[a].PB(b);
    rMap[b].PB(a);
}

bool Judge(int bak){
    for(int i=0; i<LEN; i++) Map[i].clear();
    for(int i=0; i<LEN; i++) rMap[i].clear();
     for(int i=0; i<n; i++)
        for(int j=0; j<n; j++){
            if(i == j) continue;
            int a = i*2, fa = i*2+1, b = j*2, fb = j*2+1;
            if(abs(tim[i][0]-tim[j][0]) < bak){
                addedge(a, fb);
                addedge(b, fa);        
            }
            if(abs(tim[i][0]-tim[j][1]) < bak){
                addedge(a, b);
                addedge(fb, fa);
            }
            if(abs(tim[i][1]-tim[j][0]) < bak){
                addedge(fa, fb);
                addedge(b, a);
            }
            if(abs(tim[i][1]-tim[j][1]) < bak){
                addedge(fb, a);
                addedge(fa, b);
            }
        }
    scc();
    for(int i=0; i<n*2; i+=2)
        if(sccn[i] == sccn[i+1]) return false;
    return true;
}

int main()
{
//    freopen("in.txt", "r", stdin);

    while(scanf("%d", &n)!=EOF){
        for(int i=0; i<n; i++)
            scanf("%d%d", &tim[i][0], &tim[i][1]);
        int l = 0, r = 10000000;
        while(l<r){
            int mid = (l+r+1)/2;
            if(Judge(mid))l = mid;
            else r = mid-1;
        }
        printf("%d
", l);
    }
    return 0;
}