（小米笔试）打乱字符串还原

#include <stdio.h>
#include <string.h>
int need[10][26];
void init() {
	char str[10][6] = { "ZERO","ONE","TWO","THREE","FOUR","FIVE","SIX","SEVEN","EIGHT","NINE" }; 
	int i, j;
	for (i = 0; i < 10; i++) {
		for (j = 0; j < strlen(str[i]); j++) {
			need[i][str[i][j] - 'A']++;
		}
	}
}
int find_it[10] = { 8,9,0,1,2,3,4,5,6,7 }; //搜索顺序
int ans[4000];
int dfs(int ques[],int cur) {
	int i;
	int sum = 0;
	for (i = 0; i < 26; i++)
		sum += ques[i];
	if (sum == 0)
		return cur;  //搜索成功，拥有的字符个数均为0了，返回ans的个数
	int j;
	int lp = 0;
	for (i = 0; i < 10; i++) {
		bool flag = true;
		for (j = 0; j < 26; j++) {
			if (need[find_it[i]][j] > ques[j]) {
				flag = false;
				break;
			}
		}
		if (flag) {
			for (j = 0; j < 26; j++) {//拥有的字符串都大于需要的
				ques[j] -= need[find_it[i]][j];  //减去需要的，继续搜索
			}
			ans[cur] = i;
			lp = 1;
			int k = dfs(ques, cur + 1); //如果返回了0，则表示这次搜索失败，需要继续搜索，因为是从最优的解开始搜索
			if (k != 0)                  //一旦成功就可以直接输出结果
				return k;
		}
	}
	return 0;	
}
int main() {
	int i, j;
	for (i = 0; i < 10; i++)
		for (j = 0; j < 26; j++)
			need[i][j] = 0;
	init();
	char str[10001];
	int ques[26];

	int n;
	scanf("%d", &n);
	while (n--) {
		for (i = 0; i < 4000; i++)
			ans[i] = -1;
		int num = 0;
		scanf("%s", str);
		for (i = 0; i < 26; i++) {
			ques[i] = 0;
		}
		for (i = 0; i < strlen(str); i++) {  //被打乱的字符串各个字母的个数
			ques[str[i] - 'A']++;
		}
		num = dfs(ques, 0);
		i = 0;
		while(i<num) {
			printf("%d",ans[i++]);
		}
		printf("\n");
	}
}