【USACO 2.1】Healthy Holsteins

/*
TASK: holstein
LANG: C++
URL: http://train.usaco.org/usacoprob2?a=SgkbOSkonr2&S=holstein
SOLVE: con[i][j]为食物i含有维生素j的量，ned[i]为需要的维生素i的量
bfs，用二进制保存状态
 */
#include<cstdio>
#define N 30
int v,g,ned[N],con[N][N];
int now[N];
int l,r,q[];
bool vis[];
bool ck(int s){
    //check whether state s is satisfied
    for(int i=;i<=v;i++){
        int tol=;
        for(int j=;j<g;j++)
            if((<<j)&s)
                tol+=con[j][i];
        if(tol<ned[i])return ;
    }
    return ;
}
void bfs(){
    while(l<=r){
        int k=q[l++];
        if(ck(k)){
            int num=;
            for(int i=;i<g;i++)
                if((<<i)&k)
                    num++;
            printf("%d",num);
            for(int i=;i<g;i++)
                if((<<i)&k)
                    printf(" %d",i+);
            puts("");
            return;
        }
        for(int i=;i<g;i++)
            if(!vis[k|(<<i)]){
                q[++r]=k|(<<i);
                vis[k|(<<i)]=;
            }
    }
}
int main(){
    freopen("holstein.in","r",stdin);
    freopen("holstein.out","w",stdout);
    scanf("%d",&v);
    for(int i=;i<=v;i++)
        scanf("%d",&ned[i]);
    scanf("%d",&g);
    for(int i=;i<g;i++)
        for(int j=;j<=v;j++)
            scanf("%d",&con[i][j]);
    bfs();
}