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                雙色漢諾塔問題:圓盤最初是混合顏色的從小到大排好的,現在要求根據其顏色分開到兩個柱子上分別從小到大排好。三色漢諾塔問題可與此類似,分別是排到三個柱子上。 與漢諾塔問題類似,稍作一點改動,假設柱子的編號為ABC,共有n個圓盤(n為偶數) 1.將最上面的n-1個圓盤從A移動到B上面 2.將底部那個圓盤從A移動到C 3.將B柱上的n-2個圓盤移動到A上,從而形成了第二個圖那樣的情景 因此,可以根據一個遞歸的解法來解決此問題 ![](https://box.kancloud.cn/2016-02-18_56c5c49ba376b.jpg) ~~~ //雙色漢諾塔問題 #include<iostream> using namespace std; int removeTimes = 0; void hanoiorignal(int nmovnum, char czsource, char cztemp, char czdes) { if (nmovnum == 0) { return; } else if (nmovnum == 1) { cout<<"move the "<<nmovnum<<" disk from "<<czsource <<" to "<<czdes<<endl; removeTimes++; } else { hanoiorignal(nmovnum -1, czsource, czdes, cztemp); cout<<"move the "<<nmovnum<<" disk from "<<czsource <<" to "<<czdes<<endl; removeTimes++; hanoiorignal(nmovnum -2, cztemp, czdes, czsource); } } void hanoitwocolor(int ndisk) { char szsource = 'a'; char sztemp = 'b'; char szdes = 'c'; for (int i = ndisk ; i > 0; i-= 2) { hanoiorignal(ndisk, szsource, sztemp, szdes); } cout<<"總共移動次數為:"<<removeTimes<<endl; removeTimes = 0; } void main() { int ndisknum = 0; cout<<"輸入塔上的盤碟數目,必須是2的倍數,輸入0結束"<<endl; cin>>ndisknum; while (ndisknum) { hanoitwocolor(ndisknum); cout<<"輸入塔上的盤碟數目,必須是2的倍數,輸入0結束"<<endl; cin>>ndisknum; } system("pause"); } ~~~ 輸入共有6個圓盤,輸出結果如下: ![](https://box.kancloud.cn/2016-02-18_56c5c49bb2e99.jpg)
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