CodeChef：Centeroid（树的构造）

思路&＃xff1a;假如b&＃61;2&＃xff0c;中心为A&＃xff0c;重心为B&＃xff0c;那么A-X-B&＃xff0c;为了令A和B合法&＃xff0c;必须添加节点X-X-A-X-B-(XXX)&＃xff0c;这就是b&＃61;2时需要的最少节点数&＃xff0c;B每增大1就要增加4个节点&＃xff1b;当b&＃61;0和b&＃61;1时有一些细节要处理。

# include
using namespace std;
int main()
{int t, n, b;scanf("%d",&t);while(t--){scanf("%d%d",&n,&b);if(b &＃61;&＃61; 0){if(n <3){puts("NO");continue;}puts("YES");for(int i&＃61;2; i<&＃61;n; &＃43;&＃43;i)printf("1 %d\n",i);}else if(b &＃61;&＃61; 1){if(n &＃61;&＃61; 2){printf("YES\n1 2\n");continue;}if(n &＃61;&＃61; 3){puts("NO");continue;}printf("YES\n1 2\n2 3\n3 4\n");for(int i&＃61;5; i<&＃61;n; &＃43;&＃43;i)printf("3 %d\n",i);}else{if((b-1)*4&＃43;4 > n){puts("NO");continue;}puts("YES");int p &＃61; 2*b&＃43;1;for(int i&＃61;1; i}