学步园

判断是否满流，，，人数给了100000，一看普通建图肯定不行，卡了几天了，，

看了网上说把状态相同(选择星球相同)的人合并，用二进制，最多有2^10种状态，

#include<stdio.h>
#include<string.h>
#include<queue>
#define N 100100
#define inf 0x3fffffff
using namespace std;
int dis[N],gap[10],num,head[N],start,end,ans;
struct edge
{
	int st,ed,next,flow;
}E[N*10];
void addedge(int x,int y,int flow)
{
	E[num].st=x;E[num].ed=y;E[num].flow=flow;E[num].next=head[x];head[x]=num++;
	E[num].st=y;E[num].ed=x;E[num].flow=0;   E[num].next=head[y];head[y]=num++;	
}
int dfs(int u,int minflow)
{
	if(u==end)return minflow;
	int i,v,f,flow=0,min_dis=ans-1;
	for(i=head[u];i!=-1;i=E[i].next)
	{
		v=E[i].ed;
		if(E[i].flow>0)
		{
			if(dis[v]+1==dis[u])
			{
				f=dfs(v,E[i].flow>minflow-flow?minflow-flow:E[i].flow);
				E[i].flow-=f;
				E[i^1].flow+=f;
				flow+=f;
				if(dis[start]>=ans)return flow;
				if(flow==minflow)break;
			}
			min_dis=min_dis>dis[v]?dis[v]:min_dis;
		}
	}
	if(flow==0)
	{
		if(--gap[dis[u]]==0)
			dis[start]=ans;
		dis[u]=min_dis+1;
		++gap[dis[u]];
	}
	return flow;
}
int isap()
{
	memset(dis,0,sizeof(dis));
	memset(gap,0,sizeof(gap));
	int maxflow=0;
	gap[0]=ans;
	   while(dis[start]<ans)
		   maxflow+=dfs(start,inf);
	   return maxflow;
}
int main()
{
	
    int n,m,i,j,x,bit[11],cont[1025];
	bit[0]=1;    
	for(i=1;i<=10;i++)
		bit[i]=bit[i-1]*2;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
		memset(head,-1,sizeof(head));
		num=0;
        memset(cont,0,sizeof(cont));
        for(i=1;i<=n;i++)
        {
            int temp=0;//所处的状态
            for(j=0;j<m;j++)
			{scanf("%d",&x);temp+=x*bit[j];}
            cont[temp]++;
        }
        start=0;end=1024+m+1;ans=end+1;
        for(i=0;i<1024;i++)
        {
            if(cont[i]==0)continue;
            addedge(start,i+1,cont[i]);
            for(j=0;j<10;j++)
                if(i&bit[j] )
                    addedge(i+1,1024+j+1,inf);
        }
        for(i=1;i<=m;i++)
        {
            scanf("%d",&x);
            addedge(i+1024,end,x);
        }
        if(isap()==n)printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}