• 排列 && 组合


    最近编程经常遇到需要 排列&&组合(求子集) 的问题:遂整理一下。 

    1. 数字的排列与组合(递归):O(n!),O(nC(n,k)) * O(n)

    #include <stdio.h>
    int arr[100];
    void init(int N)
    {
        for(int i = 0; i < N; ++i)
            arr[i] = i+1;
    }
    void print(int a[], int n)
    {
        for(int i = 0; i < n; ++i)
            printf("%d ", a[i]);
        printf("
    ");
    }
    
    void permutation(int arr[], int begin, int end)
    {
        if(begin == end) { print(arr, end); return; }
        for(int start = begin; start < end; ++start)
        {
            int t = arr[begin];
            arr[begin] = arr[start];
            arr[start] = t;
            permutation(arr, begin+1, end);
            arr[start] = arr[begin];
            arr[begin] = t;
        }
    }
    void combination(int endNum, int curNum, int begin, int end, int a[])
    {
        if(curNum == endNum) { print(a, endNum); return; }
        for(int i = begin; i <= end; ++i)
        {
            a[curNum] = i;
            combination(endNum, curNum+1, i+1, end, a);
        }
    }
    int main()
    {
        int N;
        scanf("%d", &N);
    
        printf("Combination: 
    ");
        int a[30] = {0};
        for(int numberOfElem = 1; numberOfElem <= N; ++numberOfElem)
            combination(numberOfElem, 0, 1, N, a);
    
        init(N);
        printf("Permutation:
    ");
    
        permutation(arr, 0, N);
        return 0;
    }
    

    shot

    2. 字符的排列组合

    #include <stdio.h>
    #include <string.h>
    char s[] = "ABC";
    
    void print(const char a[], int n)
    {
        for(int i = 0; i < n; ++i)
            printf("%c ", a[i]);
        printf("
    ");
    }
    
    void permutation(char s[], int begin, int end)
    {
        if(begin == end) { print(s, end); return; }
        for(int start = begin; start < end; ++start)
        {
            char t = s[begin];
            s[begin] = s[start];
            s[start] = t;
            permutation(s, begin+1, end);
            s[start] = s[begin];
            s[begin] = t;
        }
    }
    void combination(int endNum, int curNum, int begin, int end, char a[], const char s[])
    {
        if(curNum == endNum) { print(a, endNum); return; }
        for(char i = begin; i < end; ++i)
        {
            a[curNum] = s[i];
            combination(endNum, curNum+1, i+1, end, a, s);
        }
    }
    int main()
    {
        int N = strlen(s);
    
        printf("Combination: 
    ");
        char a[30] = {0};
        for(int members = 1; members <= N; ++members)
            combination(members, 0, 0, N, a, s);
    
        printf("Permutation:
    ");
        permutation(s, 0, N);
        return 0;
    }
    

    shot

    3.数字的组合(非递归)Θ(n2n) * O(n):

    #include <stdio.h>
    unsigned long long count = 1;
    
    int main()
    {
        int n, k, i, j, c;
        scanf("%d", &n);
        for(k = 1; k <= n; ++k)
        {
            for (i=0; i<(1<<n); i++) 
            {
                for (j=0,c=0; j<32; j++) if (i & (1<<j)) c++;
                if (c == k) 
                {
                    for (j=0;j<32; j++) if (i & (1<<j)) printf ("%i ", j+1);
                    printf ("
    ");
                    ++count;
                }
            }
        }
        printf("total: %d
    ", count);
    }

    另外:http://www.cnblogs.com/autosar/archive/2012/04/08/2437799.html 写的不错,可以学习一下。

    8月29号。

    忽然想起来小时候经常玩的快算 24.

    试着利用全排列算法,写了个快算24的小程序。效果还不错。

    思想:1、4个数字全排列,对每一个排列,从头到尾计算一遍。共 4!* 43次计算 。 2、 取出所有的两两组合,2 * C(4,2) 种方案, 共 2 * C(4,2) * 43次计算;但是利用全排列,对于每一个排列,取前两个和后两个组合整好是 2 * C(4,2) = 4!种方案。 故总计算复杂度为: 4! * 43 = 42 * 64 =  2688 次(* 2)。

    #include <iostream>
    using namespace std;
    const int v = 24;
    int A[4];
    const char ch[4] = { '+', '-', '*', '/'};
    bool has_answer = false;
    int compute(const char ch, int v1, int v2) {
    	switch(ch) {
    	case '+': return v1+v2;
    	case '-': return v1 > v2 ? v1-v2 : -10000;
    	case '*': return v1*v2;
    	default: 
    		{
    			if(v2 == 0) return -10000;
    			float tem = (float)(v1)/v2;
    			int tem2 = v1 / v2;
    			if(abs(tem-tem2) > 0) return -10000;
    			return tem2;
    		};
    	}
    }
    bool compute24() {
    	int l, m, n;
    		for(l = 0; l < 4; ++l) {
    			int v2 = compute(ch[l], A[0], A[1]);
    			for(m = 0; m < 4; ++m) {
    				int v3 = compute(ch[m], v2, A[2]);
    				for(n = 0; n < 4; ++n) {
    					if(has_answer) return true;
    					if(compute(ch[n], v3, A[3]) == v) {
    						cout << "[ [" << A[0] << ' ' << ch[l] << ' ' << A[1] << "] " << ch[m] << ' '
    							<< A[2] << " ] " << ch[n] << ' ' << A[3] << " = " << v << endl;
    						return has_answer = true;
    					}
    					else if(compute(ch[m], v2, compute(ch[n], A[2], A[3])) == v) {
    						cout << "[ " << A[0] << ' ' << ch[l] << ' ' << A[1] << " ] " << ch[m] << " [ "
    							<< A[2] << ' ' << ch[n] << ' ' << A[3] << " ] = " << v << endl;
    						return has_answer = true;
    					}
    
    				}
    			}
    		}
    	return has_answer;
    }
    
    void permutation(int begin) {
    	if(begin == 4) {
    		compute24(); 
    		return; 
    	}
    	for(int start = begin; start < 4; ++ start) {
    		int tem = A[start];
    		A[start] = A[begin];
    		A[begin] = tem;
    		permutation(begin+1);
    		A[begin] = A[start];
    		A[start] = tem;
    	}
    }
    
    int main() {
    	while(true) {
    		has_answer = false;
    		for(int k = 0; k < 4; ++k) {
    			cin >> A[k];
    		}
    		permutation(0);
    		if(!has_answer) cout << "No solution." << endl;
    	}
    	system("pause");
    	return 0;
    }
    

     准备的测试用例: {9, 10, 5, 2}, {5, 6, 7, 9}, {5, 5, 5, 5}, {7, 11, 9, 13}, {3, 7, 11, 3}, {9, 5, 6, 2}, {5, 6, 9, 11}, {10, 11, 2, 2}

    使用随机数生成:

    /* modify the function main(), as follows: */
    int main() {
    	while(true) {
    		has_answer = false;
    		srand((unsigned)time(NULL));
    		for(int k = 0; k < 4; ++k) {
    			A[k] = rand() % 14 + 1;
    			cout << A[k] << ' ';
    		}
    		cout << endl; Sleep(10000);
    		permutation(0);
    		if(!has_answer) cout << "No solution." << endl;
    		Sleep(10000);
    	}
    	system("pause");
    	return 0;
    }
    

     

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  • 原文地址:https://www.cnblogs.com/liyangguang1988/p/3703424.html
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