算法设计与应用基础：第十四周（2）

474. Ones and Zeroes

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Total Accepted: 9440
Total Submissions: 24868
Difficulty: Medium
Contributors:piy9

In the computer world, use restricted resource you have to generate maximum benefit is what we always want to pursue.
For now, suppose you are a dominator of m 

0s

 and n 

1s

 respectively.
On the other hand, there is an array with strings consisting of only 

0s

 and 

1s

.
Now your task is to find the maximum number of strings that you can form with given m 

0s

 and n 

1s

.
Each 

0

 and 

1

 can
be used at most once.
Note:

The given numbers of 

0s

 and 

1s

 will
both not exceed 

100

The size of given string array won't exceed 

600

.
解题思路：也是背包问题的变式，只不过将重量转变为0和1的个数，状态转移方程为dp[p][j]=max(dp[p-helper[i-1][0]][j-helper[i-1][1]]+1,dp[p][j]);
代码为：

int findMaxForm(vector<string>& strs, int m, int n) {
int size=strs.size();
int helper[size][2];
memset(helper,0,sizeof(helper));
for(int i=0;i<strs.size();i++)
{
int t1=0,t0=0;
for(int p=0;p<strs[i].size();p++)
{
if(strs[i][p]=='0')
t0++;
else
t1++;
}
helper[i][0]=t0;
helper[i][1]=t1;
}
int dp[m+1][n+1];
memset(dp,0,sizeof(dp));
for(int i=1;i<=size;i++)
{
for(int p=m;p>=helper[i-1][0];p--)
{

for(int j=n;j>=helper[i-1][1];j--)
{

dp[p][j]=max(dp[p-helper[i-1][0]][j-helper[i-1][1]]+1,dp[p][j]);
//else
// dp[i][p][j]=max(dp[i-1][p][j],dp[i][p][j]);
//cout<<i<<" "<<p<<" "<<j<<" "<<dp[i][p][j]<<endl;

}

}
}
return dp[m]
;

收获：空间复杂度的提高，不需要三维数组（在简单背包问题中对应于一维数组）只需要二维数组，注意一定要是简单背包问题。