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 }