给定N对括号，输出其所有的合法的组合状态，例如，，N=3，所有的合法状态为："((()))”, “(()())”, “(())()”, “()(())”, “()()()”；
解析：
还是深搜DFS的思路，深搜的过程关键在于记录已经用掉的左括号个数和右括号的个数，当用过的左括号个数小于右括号则非法；当二者个数和大于2N则非法；当二者个数相等且数目等于2N则为合法。
代码如下：
#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <set>#include <map>#include <cmath>#include <climits>#include <ctime>#include <stack>#include <queue>#include <vector>#include <algorithm>#define MAXN 1010#define RST(N)memset(N, 0, sizeof(N))using namespace std;char str[MAXN];void solve(int n, int ls, int rs){if(ls == rs && ls + rs == 2*n) {printf("%s\n", str);return ;}if(ls < rs || ls + rs >= 2*n) return ;int index = ls + rs;str[index] = '(';solve(n, ls+1, rs);str[index] = ')';solve(n, ls, rs+1);}int main(){int n;while(~scanf("%d", &n)) {solve(n, 0, 0);}return 0;}
要温暖还是怕麻烦。
