Codeforces1129E Legendary Tree

Description

Input

Output

Solution

这是一道交互题
就是有一颗n个点的树，让你在11111次询问内求出这棵树的结构
因为\(n \leqslant 500\),所哟一我们可以乱搞
我们先钦定1为根，则可以用n-1次算出以i为根的子树的节点数
然后我们将节点数排序
因为每个点只有一个父亲,我们可以通过二分那些没有父亲的点来求儿子

Code

#include <cstdio>
#include <algorithm>
#define N 1001
#define open(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout);
using namespace std;
int n,i,j,l,r,mid,tmp,cnt,len,sum,from[N],to[N],p[N];
bool bz[N];
struct node{
    int num,id;
}tree[N];
bool cmp(node x,node y){return x.num<y.num;}
int main()
{
    scanf("%d",&n);
    for (i=2;i<=n;i++)
    {
        printf("1\n1\n%d\n",n-1);
        for (j=2;j<=n;j++)
            printf("%d ",j);
        printf("\n%d\n",i);
        fflush(stdout);
        scanf("%d",&tree[i].num);
        tree[i].id=i;
    }
    tree[1].num=n;tree[1].id=1;
    sort(tree+1,tree+n+1,cmp);
    for (i=1;i<=n;i++)
    {
        if (tree[i].num==1) continue;
        tmp=1;sum=1;
        while (sum<tree[i].num)
        {
        	cnt=0;
        	for (j=1;j<i;j++)
        		if (!bz[j]) p[++cnt]=j;
        	l=1;r=cnt;
        	while (l<r)
        	{
            	mid=(l+r)/2;
            	printf("%d\n",mid-l+1);
            	for (j=l;j<=mid;j++)
               		printf("%d ",tree[p[j]].id);
            	printf("\n%d\n",1);
            	printf("%d\n%d\n",1,tree[i].id);
            	fflush(stdout);
            	scanf("%d",&tmp);
            	if (tmp)r=mid;else l=mid+1;
        	}
            bz[p[l]]=1;
            from[++len]=tree[i].id;to[len]=tree[p[l]].id;sum+=tree[p[l]].num;
		}
    }
    printf("ANSWER\n");
    for (i=1;i<=len;i++)
        printf("%d %d\n",from[i],to[i]);
    return 0;
}