For the flow of each line segment, establish a limiting formula, such as:

S _ a & lt= 26

S _ b & lt= 12

a _ b & lt= 5

....

(1) if C _ f, D _ f, D _ g and E _ g are cut off, s will not be able to connect M 1, M2, M3, M3, so C _ f, D _ f, D _ g and E _ g are one of the communication bridge combinations.

(2) It is known that the total tonnage to be transshipped is 10+8+8 = 26, in which 10 ton is M 1, 8 tons is M2 and 8 tons is M3.

(3) The sum of the four streams C _ F, D _ F, D _ G and E _ G is 10+6+5+5 = 26, so the four streams C _ F, D _ F, D _ G and E _ G are all used up, so the following four equations can be written:

(4)cf = 10，df = 6，GD = 5，eg = 5

(1) according to c_f= 10 and d_f=6, it can be known that point f will get 16 tons of goods.

According to d_g=5 and e_g=5, it can be known that point G will get 10 tons of goods.

Accordingly, it can be established that:

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How to distribute16t at point F and10t at point G to M 1 at M3 M2, and the quantity is10,8,8?

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(2) According to c_f= 10, it is known that point C will get 10 tons of goods.

According to d_f=6 and d_g=5, it is known that point D will get 1 1 ton of goods.

According to e_g=5, it is known that point E will get 5 tons of goods.

Accordingly, it can be established that:

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How to distribute the things at point S to C, D and E in the number of 10, 1 1 5?

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(1) The 10 ton goods at point G can only be delivered to M3. After M3 receives 8 tons, the remaining 2 tons can only be sent to M2. After delivery, the flow between g_M3 is 15- 10=5, and the flow between M3_M2 is 5-2=3.

(2)M 1 can only provide goods at point f, so the M 1 required 10 ton must completely pass through the connection between f_M 1.

(3) Finally, F still has 6 tons of goods to be sent to M2, which can be completely completed through the upper f_M 1 connection; You can also divert materials below 3 tons through the following f_g connection (don't forget that there are only 3 tons left between M3_M2).

Thank you, it has been solved.

The goods at 10 (1)c must come from a, so the flow of a_c is still 20- 10= 10.

(2) The 5 tons of goods at point E must come from B, so the flow of b_e is still 6-5= 1.

(3) The 1 1 ton goods at point D may come from C, B and E. 。

(3- 1) Due to the flow restriction of c_d, point C can only provide 5 tons of D at most.

(3-2) Due to the flow restriction of b_d, point B can only provide 5 tons of point D at most.

(3-3) Due to the flow restriction of b_e (refer to point 2 above), point E can only provide point D 1 ton at most.

(3-4) So the supply of goods at point D is unique: B and C are 5 tons, and E is 1 ton.

(4) Joint points 1 and 3- 1, and all points A need to provide10+5 =15t.

(5) In connection with points 2, 3-2 and 3-3, all points B need to provide 5+5+1=1ton.

(6) Combining with point 5 of point 4, we know that S will give A 15t and B 1 1tt. There are many solutions (don't forget the path of S_a_b), and the simplest one is to send it directly to A 15t through S_a and then to B65438 through S _ B..

Thank you, it has been solved.

Proof of this problem.