Sicily-1050 深度优先搜索

一．      题意

给出5个数和4则运算，看能不能算出目标值出来，如果算不出来就算出比目标值小的最大值。深搜：每一步选两个数做运算，然后算出的结果作为下一步的其中一个操作数。每一步选数有C(5,2)种，每两个数间又有5种运算结果（减法位置不同算两种）。

二．      做法：

用数组存储放进来的5各操作数，并在这个过程中用来存放中间答案，没进行一次计算操作数的个数就会减少一个。注意每一步深搜之后都要恢复原本状态。

三．      源码

 1 //

 2 //  main.cpp

 3 //  sicily-1050

 4 //

 5 //  Created by ashley on 14-10-10.

 6 //  Copyright (c) 2014年 ashley. All rights reserved.

 7 //

 8 

 9 #include <iostream>

10 using namespace std;

11 int storing[5];

12 int answer, result;

13 void depthSearch(int size)

14 {

15     for (int i = 0; i < size; i++) {

16         if (storing[i] > result && storing[i] <= answer) {

17             result = storing[i];

18         }

19     }

20     if (result == answer) {

21         return;

22     }

23     if (size == 1) {

24         return;

25     }

26     for (int i = 0; i < size - 1; i++) {

27         for (int j = i + 1; j < size; j++) {

28             int left, right;

29             left = storing[i];

30             right = storing[j];

31             storing[j] = storing[size - 1];

32             //add and decrease a number

33             storing[i] = left + right;

34             depthSearch(size - 1);

35             //subtract and decrease a number

36             storing[i] = left - right;

37             depthSearch(size - 1);

38             //subtract and decrease a number

39             storing[i] = right - left;

40             depthSearch(size - 1);

41             //multiply and decrease a number

42             storing[i] = left * right;

43             depthSearch(size - 1);

44             //divide and decrease a number

45             if (left != 0 && right % left == 0) {

46                 storing[i] = right / left;

47                 depthSearch(size - 1);

48             }

49             if (right != 0 && left % right == 0) {

50                 storing[i] = left / right;

51                 depthSearch(size - 1);

52             }

53             storing[i] = left;

54             storing[j] = right;

55         }

56     }

57 }

58 int main(int argc, const char * argv[])

59 {

60     int runs;

61     cin >> runs;

62     while (runs--) {

63         for (int i = 0; i < 5; i++) {

64             cin >> storing[i];

65         }

66         cin >> answer;

67         result = -101;

68         depthSearch(5);

69         cout << result << endl;

70     }

71     return 0;

72 }