设总结点n 则 n = n 1 + n 2 + . . . + n m + n 0 n=n_1+n_2+...+n_m+n_0 n=n1​+n2​+...+nm​+n0​ 且 n = 1 ∗ n 1 + 2 ∗ n 2 + 3 ∗ n 3 + . . . + m ∗ n m + 1 n=1*n_1+2*n_2+3*n3+...+m*n_m+1 n=1∗n1​+2∗n2​+3∗n3+...+m∗nm​+1 联立解得， n 0 = n 2 + 2 n 3 + 3 n 4 + . . . + ( m − 1 ) n m + 1 n_0=n_2+2n_3+3n_4+...+(m-1)n_m+1 n0​=n2​+2n3​+3n4​+...+(m−1)nm​+1