eg: 10x, 0<=x<=500 c(x)=1000+8x, 500<=x<=1000 3000+6x, 1000<=x<=1500 解法一: 可引入0-1变量,令z1=1,z2=1,z3=1分别表示0<=x<=500,500<=x<=1000,1000<=x<=1500,则 500z2<=x1<=500z1, 500z3<=x2<=500z2, x3<=500z3, c(x)=10x1+8x2+6x3 解法二:(更具一般性) 相关文章: 2022-02-12 2021-11-19 2022-02-08 2021-09-03 2022-02-08