学步园

dfs枚举搜索，遇到较小的路径就替换，搜索的状态为每头牛选与不选，在数目大于ans的时候直接return，

code：

/*
ID: yueqiq
LANG: C++
TASK: holstein
*/
#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <limits>
#include <vector>
#include <bitset>
#include <string>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <string.h>
#include <iostream>
#include <algorithm>
#define ls rt<<1
#define rs rt<<1|1
#define Si set<int>
#define LL long long
#define pb push_back
#define PS printf(" ")
#define Vi vector<int>
#define LN printf("\n")
#define SD(a) scanf("%d",&a)
#define PD(a) printf("%d",a)
#define SET(a,b) memset(a,b,sizeof(a))
#define FF(i,a) for(int i(0);i<(a);i++)
#define FD(i,a) for(int i(a);i>=(1);i--)
#define FOR(i,a,b) for(int i(a);i<=(b);i++)
#define FOD(i,a,b) for(int i(a);i>=(b);i--)
#define readf freopen("holstein.in","r",stdin)
#define writef freopen("holstein.out","w",stdout)
const double pi = acos(-1.0);
const int maxn = 50;
const int BigP = 99999999;
const int INF = 99999999;
const int dx[]={0,1,0,-1};
const int dy[]={1,0,-1,0};
using namespace std;
int N,M,path[maxn],ans=INF,res[maxn]={INF};
int a[maxn][maxn];
int lim[maxn],now[maxn];
bool check(int *a,int *b,int num){
    FOR(i,1,num) if(a[i]<b[i]) return false;
    return true;
}
//dfs搜索第K中饲料选或是不选
void dfs(int k,int num){
//    puts("----------------------------");
//    FOR(i,1,N) printf("%d ",now[i]);LN;
//    printf("k:%d  num:%d  ans:%d\n",k,num,ans);
//    puts("----------------------------");
    if(k>M+1) return ;
    if(num>ans) return ;
    if(check(now,lim,N)){
        bool flag=false;
        if(num==ans){
            FOR(i,0,ans-1)
                if(path[i]<res[i]){
                    flag=true;
                    break;
                }
        }else{
            ans=num;
            flag=true;
        }
        if(flag){
            FOR(i,0,ans-1) res[i]=path[i];
        }
        return;
    }
    FOR(i,1,N) now[i]+=a[k][i];
    path[num]=k;
    dfs(k+1,num+1);
    FOR(i,1,N) now[i]-=a[k][i];
    path[num]=0;
    dfs(k+1,num);
}
int main(){
    readf;
    writef;
    SD(N);
    FOR(i,1,N) SD(lim[i]);
    SD(M);
    FOR(i,1,M) FOR(j,1,N){
        SD(a[i][j]);
    }
    dfs(0,0);
    PD(ans);PS;
    FOR(i,0,ans-2){
        printf("%d ",res[i]);
    }
    printf("%d\n",res[ans-1]);
    return 0;
}