Logu P1282 Domino and p1282 Domino

Logu P1282 Domino and p1282 Domino
Description

Domino has two upper and lower squares, each of which is 1 ~ Six points. Existing

The sum of points in the upper square is S1, the sum of points in the lower square is S2, their difference is | S1-S2 |. For example in Figure 8-1, S1 = 6 + 1 + 1 + 1 = 9, S2 = 1 + 5 + 3 + 2 = 11, | S1-S2 | = 2. Each Domino card can be rotated by 180 °, so that the upper and lower squares are interchangeable. Programming uses the minimum number of rotations to minimize the number of points between the upper and lower 2 rows of a domino card.

For the example in the figure, as long as the last domino card is rotated 180 °, the difference between the two upper and lower rows is 0.

Input/Output Format

Input Format:

 

The first line of the input file is a positive integer n (1 ≤ n ≤ 1000), indicating the number of dominoes. The next n rows represent the points of n dominoes. Each line has two positive integers separated by spaces, indicating the number of points a and B in the upper and lower squares of the domino card, and 1 ≤ a, B ≤ 6.

 

Output Format:

 

The output file contains only one line and an integer. Indicates the minimum number of rotations.

 

Input and Output sample input sample #1: Copy

46 11 51 31 2

Output example #1: Copy

1

 

 

 

Ah ah

I haven't had A DP problem for A long time.

$ Dp [I] [j] $ indicates the number of the enumeration to $ I $. The difference is the minimum number of steps required by $ j $.

Next, let's take an enumeration and see if the status can be reached.

Something similar to a backpack

Note that the array may be out of bounds in this question.

So I use 1000 as 0.

 

 1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 const int MAXN=2001; 6 inline char nc() 7 { 8     static char buf[MAXN],*p1=buf,*p2=buf; 9     return p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++;10 }11 inline int read()12 {13     char c=nc();int x=0,f=1;14     while(c<'0'||c>'9'){if(c=='-')f=-1;c=nc();}15     while(c>='0'&&c<='9'){x=x*10+c-'0';c=nc();}16     return x*f;17 }18 inline int work(int x)19 {20     int ans=0;21     for(int i=1;i<x;i++)   22         if(x%i==0)  ans+=i;23     return ans;24 }25 int dp[MAXN][2*MAXN];26 int a[MAXN],b[MAXN];27 int zero=1000;28 int main()29 {30     #ifdef WIN3231     freopen("a.in","r",stdin);32     #else33     #endif34     int n=read();35     for(int i=1;i<=n;i++)    a[i]=read(),b[i]=read();36     memset(dp,0x3f,sizeof(dp));37     dp[0][1000]=0;38     for(int i=1;i<=n;i++)39     {40         for(int j=0;j<=2001;j++)41         {42             if(dp[i-1][j]<100000)43             {44                 dp[i][j+(a[i]-b[i])]=min(dp[i][j+(a[i]-b[i])],dp[i-1][j]);45                 dp[i][j+(b[i]-a[i])]=min(dp[i][j+(b[i]-a[i])],dp[i-1][j]+1);46             }47         }48     }49     int anspos=0x7ffff,ans=0x7ffff;50     for(int i=1;i<=2001;i++)51         if(dp[n][i]<=10000&&abs(i-1000)<=anspos)52             anspos=abs(i-1000);53     for(int i=1;i<=2001;i++)54         if(abs(i-1000)==anspos)55             ans=min(ans,dp[n][i]);56     printf("%d",ans);57         58     return 0;59 } 60