紫书 习题7-8 UVa 12107 (IDA*)

参考了这哥们的博客 https://blog.csdn.net/hyqsblog/article/details/46980287

 (1)atoi可以char数组转int, 头文件 cstdlib

 (2)小技巧，倒过来存是用[len-i-1]

 (3)这道题的关键在于怎么去构造这个搜索，以什么方式去搜索。这里搜索专门用两个参数来控制第几个数的第几个位置， 还有一个参数是改变的次数， 也就是深度。这里还要逆向思维，check的时候最后一个数， 可以不用递归了， 而是由前面两个数反过来推然后判断符不符合， 可以省去很多时间

#include<cstdio>
#include<cstdlib> 
#include<cstring>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;

const int MAXN = 10;
const char* word = "*0123456789";
char s[MAXN][MAXN];
int len[MAXN], maxd;

int result()
{
	int num = atoi(s[0]) * atoi(s[1]);
	char str[MAXN];
	
	REP(i, 0, len[2])
	{
		str[len[2]-i-1] = num % 10 + '0';
		num /= 10;
	}
	
	if(num != 0 || str[0] == '0') return 0;
	REP(i, 0, len[2])
		if(s[2][i] != '*' && str[i] != s[2][i])
			return 0;
			
	return 1;
}

int check(int id, int pos)
{
	if(id == 2) return result();

	int ta, tb, cnt = 0;
	if(pos == len[id] - 1) { ta = id + 1; tb = 0; }
	else { ta = id; tb = pos + 1; }
	
	char t = s[id][pos];
	if(s[id][pos] == '*')
		REP(i, 1, 11)
		{
			if(word[i] == '0' && pos == 0) continue;
			s[id][pos] = word[i];
			cnt += check(ta, tb);
			if(cnt > 1) break;
		}
	else cnt += check(ta, tb);
	
	s[id][pos] = t;
	return cnt;
}

bool dfs(int id, int pos, int d)
{
	if(d == maxd) return check(0, 0) == 1;
	if(id == 3) return false;
	
	int ta, tb;
	if(pos == len[id] - 1) { ta = id + 1; tb = 0; }
	else { ta = id; tb = pos + 1; }
	
	char t = s[id][pos];
	REP(i, 0, 11)
	{
		if(word[i] == '0' && pos == 0) continue;
		if(t == word[i]) 
		{
			if(dfs(ta, tb, d))
				return true;
		} 
		else
		{
			s[id][pos] = word[i];
			if(dfs(ta, tb, d + 1)) return true;
			s[id][pos] = t;
		}
	}
	return false;
}

int main()
{
	int kase = 0;
	while(memset(s, 0, sizeof(s)), scanf("%s%s%s", s[0], s[1], s[2]) == 3)
	{
		REP(i, 0, 3) len[i] = strlen(s[i]);
		for(maxd = 0; ; maxd++)
			if(dfs(0, 0, 0))
			{
				printf("Case %d: %s %s %s\n", ++kase, s[0], s[1], s[2]);
				break;
			}
	}
	return 0;
}