2020年数学建模B题穿越沙漠全部代码全国赛二等奖.zip

%%这个是计算第六关的第一种情况的代码 %%三个人不见面，都尽量往终点走 %% 首先到 %%Time: 2020.09.12 20:30 %%Version V2.0 Sum_first_water = 0; Sum_first_food = 0; Sum_second_water= 0; Sum_second_food = 0; Sum_third_water = 0; Sum_third_food = 0; for i = 1:1:2000 Length = Rank_scale([0,1],30,1); %%加载天气变量 [First_water1,First_food1] = RoadConsume(Length,1,1); [First_water2,First_food2] = RoadConsume(Length,2,8); [Second_water1,Second_food1] = RoadConsume(Length,1,1); [Second_water2,Second_food2] = RoadConsume(Length,2,8); Sum_first_water = Sum_first_water + 2 * First_water1 + First_water2; Sum_first_food = Sum_first_food + 2 * First_food1 + First_food2; Sum_second_water = Sum_second_water + 2 * Second_water1 + Second_water2; Sum_second_food + Sum_second_food + 2 * Second_food1 + Second_food2; [Third_water1,Third_food1] = RoadConsume(Length,1,6); Third_water2 = Length(7,1); Third_food2 = Length(7,2); [Third_water3,Third_food3] = RoadConsume(Length,8,8); Sum_third_water = Sum_third_water + Third_water1 + Third_water2 + Third_water3; Sum_third_food = Sum_third_food + Third_food1 + Third_food2 + Third_food3; end FirstWater = Sum_first_water/2000; FirstFood = Sum_first_food/2000; SecondWater= Sum_second_water/2000; SecondFood = Sum_second_food/2000; ThirdWater = Sum_third_water/2000; ThirdFood = Sum_third_food/2000; WalletMoney=30000 - ceil(FirstWater + SecondWater + ThirdWater) * 5 - ceil(FirstFood + SecondFood + ThirdFood) * 10 % % for i = 1:1:2000 % Length = Rank_scale([0,1],30,1); %%加载天气变量 % [First_water,First_food] = RoadConsume(Length,1,5); %%%前五天路上的消耗 % Sum_first_water = Sum_first_water + First_water; % Sum_first_food = Sum_first_food + First_food; % [water,food] = RoadConsume(Length,6,7); % %% % %%村庄补满，用于判断在矿上挖矿的时间 % for data = 7:1:27 % m = data % [kuang_water,kuang_food] = MinerConsume(Length,7,data); %%计算挖矿的消耗 % [mwater,mfood] = RoadConsume(Length,data+1,data+2); %%计算去村庄的路上的物资消耗 % if (water*3+food*2+kuang_water*3+kuang_food*2+mwater*3+mfood*2>=1200) % Em = m - 1 - 7 % break; % end % end % %% % %%求出挖矿的时间后，重新代入，求消耗的物资量，最终实现，最优的水和实物的搭配 % for data = 7:1:m - 1 % m = data; % [kuang_water,kuang_food] = MinerConsume(Length,7,data); %%计算挖矿的消耗 % [mwater,mfood] = RoadConsume(Length,data+1,data+2); %%计算去村庄的路上的物资消耗 % end % sumwater = sumwater + water + kuang_water + mwater; %%去矿上，挖矿，到村庄消耗 % sumfood = sumfood + food + kuang_food + mfood; % SumEm = SumEm + Em; % %% % %%再补给，在挖矿，再回家 % [Second_water,Second_food] = RoadConsume(Length,data+2,data+3); %%从村中回到矿山 % data2 = data+3 % for data = data2:1:27 % m = data % [kuang_water,kuang_food] = MinerConsume(Length,data2,data); %%计算挖矿的消耗 % [mwater,mfood] = RoadConsume(Length,data+1,data+3); %%计算村庄到终点的物资消耗 % if (Second_water*3+Second_food*2+kuang_water*3+kuang_food*2+mwater*3+mfood*2>=1200) % Second_Em = m - 1 - data2 % break; % end % end % for data = data2:1:m - 1 % m = data % [kuang_water,kuang_food] = MinerConsume(Length,data2,data); %%计算挖矿的消耗 % [mwater,mfood] = RoadConsume(Length,data+1,data+3); %%计算村庄到终点的物资消耗 % end % Secondsumwater = Secondsumwater + Second_water + kuang_water + mwater; % Secondsumfood = Secondsumfood + Second_food + kuang_food + mfood; % SumSecond_Em = SumSecond_Em + Second_Em; % end % % %% % %%通过多次累加求解得到平均值，以得到一般解 % water = sumwater/2000 % food = sumfood/2000 % Em = SumEm /2000 % First_water = Sum_first_water/2000 % First_food = Sum_first_food/2000 % Second_water= Secondsumwater/2000 % Second_food = Secondsumfood/2000 % Second_Em = SumSecond_Em/2000 % %% % %%求出最红的资金 % walletmoney = 10000 + fix(Em) * 1000 - ceil(water) * 5 - ceil(food) * 10 - ceil(First_water)*10 -ceil(First_food)*20 + fix(Second_Em) * 1000 - ceil(Second_water) * 10 - ceil(Second_food) * 20 % % %first % %%资金 % % walletmoney = % % % % 13580 % %%时间 % %T = 7 + 8 + 3 = 18 % % %second % %%%资金 % % walletmoney = % % % % 14795 % %%时间 % %T = 7 + 8 + 2 + 2 + 6 + 3 =28