题目大意不再赘述，很容易看出来是最大流，只不过人比较多，有100W个，所以需要进行压缩，可以看到m是比较小的，非常容易就能联想到2进制，所以就压缩成了1024个节点，每个结点有一个值，代表有多少人是这个状态，然后建立超级源点， 超级汇点，源点与1024个结点连线，边权为结点的值，然后每个源点再与m个星球进行连线，边权为无限大，然后每个星球再与汇点进行连线，值为该星球的容纳量。
/*
ID: CUGB-wwj
PROG:
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define INF 1111111111
#define MAXN 100005
#define MAXM 444444
#define PI acos(-1.0)
using namespace std;
struct node
{int ver; // vertexint cap; // capacityint flow; // current flow in this arcint next, rev;
}edge[MAXM];
int dist[MAXN], numbs[MAXN], src, des, n;
int head[MAXN], e;
void add(int x, int y, int c)
{ //e记录边的总数edge[e].ver = y;edge[e].cap = c;edge[e].flow = 0;edge[e].rev = e + 1; //反向边在edge中的下标位置edge[e].next = head[x]; //记录以x为起点的上一条边在edge中的下标位置head[x] = e++; //以x为起点的边的位置//反向边edge[e].ver = x;edge[e].cap = 0; //反向边的初始网络流为0edge[e].flow = 0;edge[e].rev = e - 1;edge[e].next = head[y];head[y] = e++;
}
void rev_BFS()
{int Q[MAXN], qhead = 0, qtail = 0;for(int i = 1; i <= n; ++i){dist[i] = MAXN;numbs[i] = 0;}Q[qtail++] = des;dist[des] = 0;numbs[0] = 1;while(qhead != qtail){int v = Q[qhead++];for(int i = head[v]; i != -1; i = edge[i].next){if(edge[edge[i].rev].cap == 0 || dist[edge[i].ver] < MAXN)continue;dist[edge[i].ver] = dist[v] + 1;++numbs[dist[edge[i].ver]];Q[qtail++] = edge[i].ver;}}
}
void init()
{e = 0;memset(head, -1, sizeof(head));
}
int maxflow()
{int u, totalflow = 0;int Curhead[MAXN], revpath[MAXN];for(int i = 1; i <= n; ++i)Curhead[i] = head[i];u = src;while(dist[src] < n){if(u == des) // find an augmenting path{int augflow = INF;for(int i = src; i != des; i = edge[Curhead[i]].ver)augflow = min(augflow, edge[Curhead[i]].cap);for(int i = src; i != des; i = edge[Curhead[i]].ver){edge[Curhead[i]].cap -= augflow;edge[edge[Curhead[i]].rev].cap += augflow;edge[Curhead[i]].flow += augflow;edge[edge[Curhead[i]].rev].flow -= augflow;}totalflow += augflow;u = src;}int i;for(i = Curhead[u]; i != -1; i = edge[i].next)if(edge[i].cap > 0 && dist[u] == dist[edge[i].ver] + 1)break;if(i != -1) // find an admissible arc, then Advance{Curhead[u] = i;revpath[edge[i].ver] = edge[i].rev;u = edge[i].ver;}else // no admissible arc, then relabel this vertex{if(0 == (--numbs[dist[u]]))break; // GAP cut, Important!Curhead[u] = head[u];int mindist = n;for(int j = head[u]; j != -1; j = edge[j].next)if(edge[j].cap > 0)mindist = min(mindist, dist[edge[j].ver]);dist[u] = mindist + 1;++numbs[dist[u]];if(u != src)u = edge[revpath[u]].ver; // Backtrack}}return totalflow;
}
int in()
{char ch;int a = 0;while((ch = getchar()) == ' ' || ch == '\n');a += ch - '0';while((ch = getchar()) != ' ' && ch != '\n'){a *= 10;a += ch - '0';}return a;
}
int peo[1111];
int val[55];
int main()
{int num, m;int u, v;while(scanf("%d%d", &num, &m) != EOF){init();memset(peo, 0, sizeof(peo));src = 1;des = 1 + (1 << m) + m + 1;n = des;for(int i = 0; i < num; i++){int t = 0;for(int j = 0; j < m; j++){u = in();if(u) t += (1 << j);}peo[t]++;}for(int i = 0; i < m; i++){scanf("%d", &val[i]);add(i + (1 << m) + 2, des, val[i]);}for(int i = 0; i < (1 << m); i ++){if(peo[i] == 0) continue;add(src, i + 2, peo[i]);for(int j = 0; j < m; j++){int k = 1 << j;if(i & k) add(i + 2, j + (1 << m) + 2, INF);}}rev_BFS();int ans = maxflow();if(ans >= num) printf("YES\n");else printf("NO\n");}return 0;
}