例题：枚举排列

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枚举排列常见的方法有两种：一种是递归枚举，而是用STL中的next_permutation

【1】生成可重集的排列

#include<stdio.h>#include<string.h>#include<stdlib.h>int p[20];int cmp(const void *a, const void *b){return *(int *)a - *(int *)b;}void print_permutation(int n, int *a, int cur){if (cur == n){for (int i = 0; i < n; i++)printf("%d ", a[i]);printf("\n");}else{for (int i = 0; i < n; i++){if (!i || p[i] != p[i - 1])//防止重复的数多次输出，需要先排序{int c1 = 0, c2 = 0;//为了顺利输出存在相同的情况for (int j = 0; j < cur; j++)if (a[j] == p[i])c1++;//寻找现排列中与原排列相同的个数c1for (int j = 0; j < n; j++)if (p[j] == p[i])c2++;//原排列中自身重复的个数c2if (c1 < c2)//c1<c2即可递归{a[cur] = p[i];print_permutation(n, a, cur + 1);}}}}}int main(){int a[20];int n;scanf("%d", &n);for (int i = 0; i < n; i++)scanf("%d", &p[i]);qsort(p, n, sizeof(p[0]), cmp);print_permutation(n, a, 0);return 0;}

【2】下一个排列

#include<cstdio>#include<algorithm>using namespace std;int main(){int n, p[10];scanf("%d", &n);for (int i = 0; i < n; i++)scanf("%d", &p[i]);sort(p, p + n);do {for (int i = 0; i < n; i++)printf("%d ", p[i]);printf("\n");} while (next_permutation(p, p + n));//直接while不会输出自身（输入的那个序列）return 0;}

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