URAL 1997 Those are not the droids you're looking for

二分图的最大匹配。

每一个$0$与$1$配对，只建立满足时差大于等于$a$或者小于等于$b$的边，如果二分图最大匹配等于$n/2$，那么有解，遍历每一条边输出答案，否则无解。

#include<map>
#include<set>
#include<ctime>
#include<cmath>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define ms(x,y) memset(x,y,sizeof(x))
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define per(i,j,k) for(int i=j;i>=k;i--)
#define loop(i,j,k) for (int i=j;i!=-1;i=k[i])
#define inone(x) scanf("%d",&x)
#define intwo(x,y) scanf("%d%d",&x,&y)
#define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define infou(x,y,z,p) scanf("%d%d%d%d",&x,&y,&z,&p)
#define lson x<<1,l,mid
#define rson x<<1|1,mid+1,r
#define mp(i,j) make_pair(i,j)
#define ft first
#define sd second
typedef long long LL;
typedef pair<int, int> pii;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 1e6 + 10;
const int M = 1e4 + 1;
const double eps = 1e-10;

int a,b,n;
int T[1010],f[1010];

const int maxn = 1010 + 10;
struct Edge
{
    int from, to, cap, flow;
    Edge(int u, int v, int c, int f) :from(u), to(v), cap(c), flow(f){}
};
vector<Edge>edges;
vector<int>G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn];
int s, t;

void init()
{
    for (int i = 0; i < maxn; i++)
        G[i].clear();
    edges.clear();
}
void AddEdge(int from, int to, int cap)
{
    edges.push_back(Edge(from, to, cap, 0));
    edges.push_back(Edge(to, from, 0, 0));
    int w = edges.size();
    G[from].push_back(w - 2);
    G[to].push_back(w - 1);
}
bool BFS()
{
    memset(vis, 0, sizeof(vis));
    queue<int>Q;
    Q.push(s);
    d[s] = 0;
    vis[s] = 1;
    while (!Q.empty())
    {
        int x = Q.front();
        Q.pop();
        for (int i = 0; i<G[x].size(); i++)
        {
            Edge e = edges[G[x][i]];
            if (!vis[e.to] && e.cap>e.flow)
            {
                vis[e.to] = 1;
                d[e.to] = d[x] + 1;
                Q.push(e.to);
            }
        }
    }
    return vis[t];
}
int DFS(int x, int a)
{
    if (x == t || a == 0)
        return a;
    int flow = 0, f;
    for (int &i = cur[x]; i<G[x].size(); i++)
    {
        Edge e = edges[G[x][i]];
        if (d[x]+1 == d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0)
        {
            edges[G[x][i]].flow+=f;
            edges[G[x][i] ^ 1].flow-=f;
            flow+=f;
            a-=f;
            if(a==0) break;
        }
    }
    if(!flow) d[x] = -1;
    return flow;
}
int dinic(int s, int t)
{
    int flow = 0;
    while (BFS())
    {
        memset(cur, 0, sizeof(cur));
        flow += DFS(s, INF);
    }
    return flow;
}

int main()
{
    while(~scanf("%d%d",&a,&b))
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d%d",&T[i],&f[i]);

        init();

        for(int i=1;i<=n;i++)
        {
            if(f[i]==1) continue;
            for(int j=i+1;j<=n;j++)
            {
                if(f[j]==0) continue;
                if(T[j]-T[i]>=a||T[j]-T[i]<=b)
                {
                    AddEdge(i,j,1);
                }
            }
        }

        s=0,t=n+1;

        for(int i=1;i<=n;i++)
        {
            if(f[i]==0) AddEdge(s,i,1);
            else AddEdge(i,t,1);
        }

        int M = dinic(s,t);

        if(M<n/2)
        {
            printf("Liar
");
        }
        else
        {
            printf("No reason
");
            for(int i=0;i<edges.size();i++)
            {
                if(edges[i].flow!=1) continue;
                if(edges[i].from!=s&&edges[i].to!=t)
                    printf("%d %d
",T[edges[i].from],T[edges[i].to]);
            }
        }

    }
    return 0;
}