Barber of the Army of Mages

Barber of the Army of Mages

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题意：n个顾客，每个顾客要理两次发，而且只会在[ti, ti+si-1]的时间内出现，发廊每个时间只能接待k个客人，问能否满足所有客人。可以就输出每个时间段招待的顾客
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解法：网络流，顾客拆分成两个，跑最大流即可、
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队友代码：

#include<cstdio>
#include<vector>
#include<cstring>
#include<queue>
using namespace std;

#define repi(i,A,B) for(int i=A;i<=B;i++)
#define rep(i,A,B) for(i=A;i<=B;i++)
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))

const int MAXN=2500;
const int MAXM=1000010;
const int INF=0x3f3f3f3f;
struct Edge
{
    int to,next,cap,flow;
}edge[MAXM];
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
int ar[MAXN],la[MAXN];
int n,k;


void swap(int &a,int &b)
{
    int c=a;
    a=b;b=c;
}

void init()
{
    tol=0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int w,int rw=0)
{
    edge[tol].to=v; edge[tol].cap=w; edge[tol].flow=0;
    edge[tol].next=head[u]; head[u]=tol++;
    edge[tol].to=u; edge[tol].cap=rw; edge[tol].flow=0;
    edge[tol].next=head[v]; head[v]=tol++;
}

int Q[MAXN];
void BFS(int start,int end)
{
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0]=1;
    int front=0, rear=0;
    dep[end]=0;
    Q[rear++]=end;
    while(front!=rear)
    {
        int u=Q[front++];
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if(dep[v]!=-1) continue;
            Q[rear++]=v;
            dep[v]=dep[u]+1;
            gap[dep[v]]++;
        }
    }
}
int S[MAXN];
int sap(int start,int end,int N)
{
    BFS(start,end);
    memcpy(cur,head,sizeof(head));
    int top=0;
    int u=start;
    int ans=0;
    while(dep[start]<N)
    {
        if(u==end)
        {
            int Min=INF;
            int inser;
            for(int i=0;i<top;i++)
            {
                if(Min>edge[S[i]].cap-edge[S[i]].flow)
                {
                    Min=edge[S[i]].cap-edge[S[i]].flow;
                    inser=i;
                }
            }
            for(int i=0;i<top;i++)
            {
                edge[S[i]].flow+=Min;
                edge[S[i]^1].flow-=Min;
            }
            ans+=Min;
            top=inser;
            u=edge[S[top]^1].to;
            continue;
        }
        bool flag=false;
        int v;
        for(int i=cur[u];i!=-1;i=edge[i].next)
        {
            v=edge[i].to;
            if(edge[i].cap-edge[i].flow && dep[v]+1==dep[u])
            {
                flag=true;
                cur[u]=i;
                break;
            }
        }
        if(flag)
        {
            S[top++]=cur[u];
            u=v;
            continue;
        }
        int Min=N;
        for(int i=head[u];i!=-1;i=edge[i].next)
            if(edge[i].cap-edge[i].flow && dep[edge[i].to]<Min)
            {
                Min=dep[edge[i].to];
                cur[u]=i;
            }
        gap[dep[u]]--;
        if(!gap[dep[u]]) return ans;
        dep[u]=Min+1;
        gap[dep[u]]++;
        if(u!=start) u=edge[S[--top]^1].to;
    }
    return ans;
}


int main()
{

    int maxx=0;
    scanf("%d%d",&n,&k);
    init();
    repi(i,1,n)
    {
        scanf("%d%d",ar+i,la+i);
        maxx=max(maxx,ar[i]+la[i]-1);
    }
    //printf("%d\n",maxx);
    repi(i,0,maxx) addedge(2150,i,k);
    repi(i,1,n)
    {
        repi(j,0,maxx)
        {
            if(j>=ar[i] && j<=ar[i]+la[i]-1)
            {
                addedge(j,i+2000,1);
               // printf("%d %d\n",j,i+2000);
            }
        }
        addedge(i+2000,2200,2);
    }
    int mf=sap(2150,2200,maxx+n+2);
    if(mf!=2*n)
    {
        printf("No\n");
        return 0;
    }
    else
    {
        printf("Yes\n");
        repi(i,1,n)
        {
            int ans[3],cnt=0;
            for(int j=head[i+2000];j!=-1;j=edge[j].next)
            {
                if(edge[j].to!=2200 && edge[j].flow==-1)
                {
                    ans[++cnt]=edge[j].to;
                }
            }
            if(ans[1]>ans[2]) swap(ans[1],ans[2]);

            printf("%d %d\n",ans[1],ans[2]);
        }

    }
    return 0;
}