题目链接：
URAL 1997 Those are not the droids you’re looking for
题意：
给出一间酒吧的n个进出记录，判断进出这间酒吧的所有人是不是只属于这两种：
①：待在酒吧的时间不少于a
②：待在酒吧的时间至多为b
如果存在一种方案使得根据这n个进出记录的到的所有人都属于这两种人输出每个人的进出时间。否则输出”Liar”。
分析：
二分图匹配。
进入时间和出去时间的匹配。如果满足上述条件，就增加一条增广轨。判断最大匹配是否恰好为完全匹配。
特别注意：出去时间一定是大于进入时间的！
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <climits>
#include <cmath>
#include <ctime>
#include <cassert>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;
const int MAX_N = 1010;int n, ids, ide, total;
ll a, b, s[MAX_N], e[MAX_N];
int matchx[MAX_N], matchy[MAX_N], vis[MAX_N], head[MAX_N];struct Edge{int to, next;Edge() {}Edge(int _to, int _next) :to(_to), next(_next) {}
}edge[MAX_N * MAX_N];void AddEdge(int from, int to)
{edge[total].to = to;edge[total].next = head[from];head[from] = total++;
}bool dfs(int u)
{for(int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].to;if(vis[v]) continue;vis[v] = 1;if(matchy[v] == -1 || dfs(matchy[v])){matchx[u] = v;matchy[v] = u;return true;}}return false;
}int Hungary()
{memset(matchx, -1, sizeof(matchx));memset(matchy, -1, sizeof(matchy));int res = 0;for(int i = 0; i < ids; i++){memset(vis, 0, sizeof(vis));if(dfs(i)) res++;}return res;
}int main()
{freopen("H.in", "r", stdin);IOS;while(cin >> a >> b){cin >> n;memset(head, -1, sizeof(head));total = ids = ide = 0;for(int i = 0; i < n; i++){ll t;int id;cin >> t >> id;if(id == 0) s[ids++] = t;else e[ide++] = t;}for(int i = 0; i < ids; i++){for(int j = 0; j < ide; j++){if(e[j] <= s[i]) continue;if(s[i] + b >= e[j] || s[i] + a <= e[j]){AddEdge(i, j);}}}int ans = Hungary();//cout << ans << endl;if(ids != ide && n & 1 || ans < n / 2) cout << "Liar" << endl;else {cout << "No reason" << endl;for(int i = 0; i < ids; i++){cout << s[i] << " " << e[matchx[i]] << endl;}}}return 0;
}