SCU 4443 Range Query

二分图最大匹配，枚举。

可以计算出每一个位置可以放哪些数字，每个数字可以放在哪些位置，这样就可以建二分图了。

如果二分图最大匹配不到$n$，则无解。否则构造字典序最小的解，可以枚举每一位放什么数字，然后再判断是否有解。

#include<bits/stdc++.h>
using namespace std;

const int maxn=+;
int n,m1,m2;
int pL[],pR[],nL[],nR[];
int f,ans[],cun[],u[][];

const int INF = 0x7FFFFFFF;
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 = ; i < maxn; i++) G[i].clear();
    edges.clear();
}
void Addedge(int from, int to, int cap)
{
    edges.push_back(Edge(from, to, cap, ));
    edges.push_back(Edge(to, from, , ));
    int w = edges.size();
    G[from].push_back(w - );
    G[to].push_back(w - );
}
bool BFS()
{
    memset(vis, , sizeof(vis));
    queue<int>Q;
    Q.push(s);
    d[s] = ;
    vis[s] = ;
    while (!Q.empty())
    {
        int x = Q.front();
        Q.pop();
        for (int i = ; i<G[x].size(); i++)
        {
            Edge e = edges[G[x][i]];
            if (!vis[e.to] && e.cap>e.flow)
            {
                vis[e.to] = ;
                d[e.to] = d[x] + ;
                Q.push(e.to);
            }
        }
    }
    return vis[t];
}
int DFS(int x, int a)
{
    if (x == t || a == )
        return a;
    int flow = , f;
    for (int &i = cur[x]; i<G[x].size(); i++)
    {
        Edge e = edges[G[x][i]];
        if (d[x]+ == d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>)
        {
            edges[G[x][i]].flow+=f;
            edges[G[x][i] ^ ].flow-=f;
            flow+=f;
            a-=f;
            if(a==) break;
        }
    }
    if(!flow) d[x] = -;
    return flow;
}
int dinic(int s, int t)
{
    int flow = ;
    while (BFS())
    {
        memset(cur, , sizeof(cur));
        flow += DFS(s, INF);
    }
    return flow;
}

int main()
{
    while(~scanf("%d%d%d",&n,&m1,&m2))
    {
        f=; memset(u,,sizeof u);
        for(int i=;i<=n;i++) nL[i]=pL[i]=,nR[i]=pR[i]=n;

        for(int i=;i<=m1;i++)
        {
            int a,b,c; scanf("%d%d%d",&a,&b,&c);
            for(int j=a;j<=b;j++) pL[j]=max(c,pL[j]);
            nL[c] = max(nL[c],a);
            nR[c] = min(nR[c],b);
        }

        for(int i=;i<=m2;i++)
        {
            int a,b,c; scanf("%d%d%d",&a,&b,&c);
            for(int j=a;j<=b;j++) pR[j]=min(c,pR[j]);
            nL[c] = max(nL[c],a);
            nR[c] = min(nR[c],b);
        }

        for(int i=;i<=n;i++)
        {
            if(nL[i]>nR[i]) f=;
            if(pL[i]>pR[i]) f=;
        }

        if(f==)
        {
            printf("-1\n");
            continue;
        }

        init();

        s=, t=*n+;

        for(int i=;i<=n;i++) Addedge(s,i,), Addedge(i+n,t,);

        for(int i=;i<=n;i++)
            for(int j=pL[i];j<=pR[i];j++)
                if(nL[j]<=i&&i<=nR[j]) Addedge(i,j+n,), u[i][j]=;

        int pi = dinic(s,t);

        if(pi!=n)
        {
            printf("-1\n");
            continue;
        }

        memset(cun,,sizeof cun);
        for(int pos=;pos<=n;pos++)
        {
            for(int num=;num<=n;num++)
            {
                if(cun[num]) continue;

                if(u[pos][num]==) continue;

                init();
                s=, t=*n+;
                for(int i=;i<=n;i++)
                {
                    if(i>pos) Addedge(s,i,);
                    if(cun[i]==&&i!=num) Addedge(i+n,t,);
                }

                for(int i=pos+;i<=n;i++)
                    for(int j=pL[i];j<=pR[i];j++)
                        if(nL[j]<=i&&i<=nR[j])
                            if(cun[j]==&&j!=num) Addedge(i,j+n,);

                pi = dinic(s,t);
                if(pi==n-pos)
                {
                    ans[pos]=num;
                    cun[num]=;
                    break;
                }
            }
        }

        for(int i=;i<=n;i++)
        {
            printf("%d",ans[i]);
            if(i<n) printf(" ");
            else printf("\n");
        }
    }
    return ;
}