1501. Sense of Beauty
Time limit: 0.5 second
Memory limit: 64 MB
The owner of a casino for New Russians has a very refined sense of beauty. For example, after a game there remain two piles with the same number of cards on the table, and the owner likes the cards
to be arranged into two piles according to the color: one pile with red cards and the other with black cards. Of course, this is done not by the owner himself, but by a croupier. The owner just likes to watch the process. The croupier takes a card from the
top of one of the initial piles and puts it into one of the new piles; this is repeated until all the cards from the initial piles are transferred. The owner doesn‘t like it if one of the resulting piles grows faster than the other. At each moment the resulting
piles must not differ in size by more than one card; a bigger difference would contradict the owner‘s sense of beauty. Help the croupier to arrange the cards according to the tastes of his owner.
Input
The first line of the input contains the number N of cards in each of the piles (4 ≤ N ≤ 1000). Each of the next two lines contains N digits 0 or 1 describing the piles: 1
denotes a red-suit card and 0 denotes a black-suit card. The cards in a pile are described from the top to the bottom. There are in total N red and N black cards in the two piles.
Output
Output a line containing 2N digits 1 or 2, which describes the process of transferring the cards. Each number shows the number of the pile from which a card is taken. If it is impossible to
perform this task according to the given rules, output "Impossible".
Samples
| input | output |
|---|---|
4 0011 0110 |
22121112 |
4 1100 1100 |
Impossible |
题意：
给你两堆牌，牌的颜色只有红色或者黑色。 然后从两堆牌的牌顶来抽牌，每次抽可以选择两堆中的一堆。每次抽完，所得到的牌，红牌和黑牌数量相差必须不超过1。
做法：
因为一共各1000张牌，所以可以dp记忆化搜索。dp[i][j]代表 在第一堆牌抽了i张，第二堆牌抽了j张的情况下， 有没有不违反规则 达到这状态的方法。如果有 dp[i][j]会等于0，1，2，0表示当前多了一个黑牌，1表示当前红黑牌一样多，2表示当前红牌多一张。-2表示没达到这种状态的方法。
然后就是几种转移的方法，都要在dfs（i-1，j）或者 dfs(j,i-1) 可以达到的情况下 才能转移。
int dp[1100][1100];//
int n;
int num1[1100];
int num2[1100];
int bu[1100];
int dfs(int i,int j)
{
if(~dp[i][j])//不是-1，就返回已经有的值， 记忆化搜索，不然会超时
return dp[i][j];
if(i==0&&j==0)
return dp[i][j]=1;//0 0多 1平 2 1多
if(i!=0)
{
if(dfs(i-1,j)==0&&num1[i]==1)//这里表示，上一个状态 红黑牌中 红牌多，所以这次必须抽黑牌才能转移
{
bu[i+j]=1;
return dp[i][j]=1;
}
if(dfs(i-1,j)==1)
{
if(num1[i]==1)
{
bu[i+j]=1;
return dp[i][j]=2;
}
if(num1[i]==0)
{
bu[i+j]=1;
return dp[i][j]=0;
}
}
if(dfs(i-1,j)==2&&num1[i]==0)
{
bu[i+j]=1;
return dp[i][j]=1;
}
}
if(j!=0)
{
if(dfs(i,j-1)==0&&num2[j]==1)
{
bu[i+j]=2;
return dp[i][j]=1;
}
if(dfs(i,j-1)==1)
{
if(num2[j]==1)
{
bu[i+j]=2;
return dp[i][j]=2;
}
if(num2[j]==0)
{
bu[i+j]=2;
return dp[i][j]=0;
}
}
if(dfs(i,j-1)==2&&num2[j]==0)
{
bu[i+j]=2;
return dp[i][j]=1;
}
}
return dp[i][j]=-2;
}
char n1[1100];
char n2[1100];
int main()
{
while(scanf("%d",&n)!=EOF)
{
scanf("%s%s",n1,n2);
memset(dp,-1,sizeof dp);
for(int i=1;i<=n;i++)
{
num1[i]=n1[i-1]-'0';
num2[i]=n2[i-1]-'0';
}
if(dfs(n,n)!=-2)
{
for(int i=1;i<=2*n;i++)
printf("%d",bu[i]);
puts("");
}
else
puts("Impossible");
}
return 0;
}