HDU 2234 Untitled I

Untitled itime limit:10000/10000ms (Java/other) Memory limit:32768/32768k (Java/other) total submission (s): 2 Accepted Su Bmission (s): 1Problem description One day robot little A is playing a simple intellectual game, this game is like this, in a 4*4 matrix of 4 1, 4 2, 4 3 and 4 4 different things, respectively, Each step of small a can move the 4 number of the same row to the left or to the right one step or the same column of 4 numbers up or down one step (1,2,3,4 to the left after the 2,3,4,1), Little a now wants to know that the minimum number of moves can be the same as the 4 numbers on each line of the matrix or the 4 numbers on each column. But little a does not want to go too many steps, he just know the minimum number of steps is equal to 5 steps, the output is the correct number of steps, otherwise output-1.
Input first enters an integer t, which indicates that there is a T group of data. For each set of data, enter 4 rows, and each row of 4 columns represents the matrix.
Output a positive integer for each set of input outputs represents the minimum number of moving steps, which is greater than 5 for output-1.
Sample Input

21 2 3 41 2 3 41 2 3 42 3 4 14 1 1 11 2 2 22 3 3 33 4 4 4

Sample Output

11

Sourcehdoj Summer Exercise (2)-Hold by Captain Xu

Change four point states at a time

The estimated four changes are correct at least the number of moves required is (ans+3)/4;

#include <iostream> #include <cstdio> #include <queue> #include <cstring> #include <string    > #define INF 1<<30using namespace std;int get_h (int b[4][4]) {int ans=inf,tmp=0;        for (int i=0;i<4;i++) {bool flag[5];        int cnt=4;        memset (flag,false,sizeof (flag));                for (int j=0;j<4;j++) if (!flag[b[i][j]]) {cnt--;            Flag[b[i][j]]=true;    } tmp+=3-cnt;    } ans=min (Tmp,ans);    Tmp=0;        for (int j=0;j<4;j++) {bool flag[5];        int cnt=4;        memset (flag,false,sizeof (flag));                for (int i=0;i<4;i++) if (!flag[b[i][j]]) {cnt--;            Flag[b[i][j]]=true;    } tmp+=3-cnt;    } ans=min (Tmp,ans); Return (ANS+3)/4;}    BOOL Dfs (int len,int a[4][4],int kind,int kind1) {if (Get_h (a) >len) return false;    if (len==0) return true;    int aa[4][4]; for (int i=0; i<4; i++) {if (kind==i&&kind1==2) {; } else {for (int j=0; j<4; j + +) for (int k=0; k<4; k++) AA [j]            [K]=a[j][k];                for (int j=0; j<4; J + +) if (j) aa[i][j]=a[i][j-1];            else aa[i][j]=a[i][3];        if (Dfs (len-1,aa,i,1)) return true;        } if (kind==i&&kind1==1); else {for (int j=0; j<4; j + +) for (int k=0; k<4; k++) aa[j][k]=a[            J][K];                for (int j=0; j<4; J + +) if (j!=3) aa[i][j]=a[i][j+1];            else aa[i][j]=a[i][0];        if (Dfs (len-1,aa,i,2)) return true;        } if (kind==i&&kind1==3) {; } else {for (int j=0; j<4; j + +) for (int k=0; k<4; k++) AA [j]            [K]=a[j][k];       for (int j=0; j<4; J + +) if (j!=3) aa[j][i]=a[j+1][i];         else Aa[j][i]=a[0][i];        if (Dfs (len-1,aa,i,4)) return true;        } if (kind==i&&kind1==4) {; }else {for (int j=0; j<4; j + +) for (int k=0; k<4; k++) aa[j][k]=a [j]            [K];                for (int j=0; j<4; J + +) if (j) Aa[j][i]=a[j-1][i];            else Aa[j][i]=a[3][i];        if (Dfs (len-1,aa,i,3)) return true; }} return false;}    int main () {int t;    scanf ("%d", &t);        while (t--) {int a[4][4];        for (int. i=0; i<4; i++) {for (int j=0; j<4; j + +) scanf ("%d", &a[i][j]);        } int len; For (Len=get_h (a); len<=5; len++) {if (Dfs (len,a,-1,-1)) {printf ("%d\n", Len                );            Break    }} if (len>5) printf (" -1\n"); }}

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HDU 2234 Untitled I