码迷,mamicode.com
首页 > 其他好文 > 详细

hdu 5755(GAuss 消元)

时间:2016-09-11 23:07:39      阅读:339      评论:0      收藏:0      [点我收藏+]

标签:

Gambler Bo

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1152    Accepted Submission(s): 471
Special Judge


Problem Description
Gambler Bo is very proficient in a matrix game.

You have a N×M matrix, every cell has a value in {0,1,2}.

In this game, you can choose a cell in the matrix, plus 2 to this cell, and plus 1 to all the adjacent cells.

for example, you choose the cell (x,y), the value of (x,y) will be plused 2, and the value of (x1,y)(x+1,y)(x,y1)(x,y+1) will be plused 1.

if you choose the cell (1,2), the cell (1,2) will be plused 2, and the cell (2,2)(1,1)(1,3) will be plused 1, the cell (0,2) won‘t be changed because it‘s out of the matrix.

If the values of some cells is exceed 2, then these values will be modulo 3.

Gambler Bo gives you such a matrix, your task is making all value of this matrix to 0 by doing above operations no more than 2NM times.
 

 

Input
First line, an integer T. There are T test cases.

In each test, first line is two integers N,M, and following N lines describe the matrix of this test case.

T10,1N,M30, the matrix is random and guarantee that there is at least one operation solution.
 

 

Output
For each test, first line contains an integer num(0num2NM) describing the operation times.

Following num lines, each line contains two integers x,y(1xN,1yM) describing the operation cell.

The answer may not be unique, you can output any one.
 

 

Sample Input
2 2 3 2 1 2 0 2 0 3 3 1 0 1 0 1 0 1 0 1
 

 

Sample Output
1 1 2 5 1 1 1 3 2 2 3 1 3 3
 

 

Author
绍兴一中
 

 

Source
 

 

Recommend
wange2014

 


gauss消元的mod 3 版本,把n*m个格子上的操作全部变成列向量,共n*m个,每个列向量有n*m个元素,做一遍gauss消元就能解出。

另:优化后因为第i行的状态可以由i+1行确定,所以能建立m个方程,每个方程 M 个变量进行高斯消元,解出解后代回去得到每个元素应该被操作的次数。详见

http://www.cnblogs.com/naturepengchen/articles/5711133.html

我的裸gauss消元代码如下:

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<cmath>
  6 #define clr(x) memset(x,0,sizeof(x))
  7 #define clrdown(x) memset(x,-1,sizeof(x))
  8 #define maxn 910
  9 using namespace std;
 10 int A[maxn][maxn];
 11 int free_x[maxn];
 12 int x[maxn];
 13 int mov[4][2]={0,1,0,-1,1,0,-1,0};
 14 void init(int n,int m);
 15 void gauss(int n,int m);
 16 void print(int n,int m);
 17 int exgcd(int a,int b,int &x,int &y);
 18 int lcm(int a,int b);
 19 int gcd(int a,int b);
 20 int main()
 21 {
 22     int T,p,num,n,m;
 23     scanf("%d",&T);
 24     while(T--)
 25     {
 26         scanf("%d%d",&n,&m);
 27         init(n,m);
 28         gauss(n*m,n*m);
 29         print(n,m);
 30     }
 31     return 0;
 32 }
 33 void print(int n,int m)
 34 {
 35     int sum=0;
 36     for(int i=0;i<n*m;i++)
 37         sum+=x[i];
 38     printf("%d\n",sum);
 39     for(int i=0;i<n;i++)
 40         for(int j=0;j<m;j++)
 41             while(x[i*m+j]>0)
 42             {
 43                 printf("%d %d\n",i+1,j+1);
 44                 x[i*m+j]--;
 45             }
 46     return ;
 47 }
 48 void init(int n,int m)
 49 {
 50     int t;
 51     clr(A);
 52     for(int i=0;i<n;i++)
 53         for(int j=0;j<m;j++)
 54         {
 55             t=i*m+j;
 56             A[t][t]=2;
 57             for(int k=0;k<4;k++)
 58                 if(i+mov[k][0]>=0 && i+mov[k][0]<n && j+mov[k][1]>=0 && j+mov[k][1]<m)
 59                     A[(i+mov[k][0])*m+j+mov[k][1]][t]=1;
 60         }
 61     t=n*m;
 62     for(int i=0;i<n;i++)
 63         for(int j=0;j<m;j++)
 64         {
 65             scanf("%d",&A[i*m+j][t]);
 66             A[i*m+j][t]=(3-A[i*m+j][t])%3;
 67         }
 68     clrdown(x);
 69     clr(free_x);
 70 }
 71 void gauss(int n,int m)
 72 {
 73 //    for(int i=0;i<n;i++)
 74 //    {
 75 //       for(int j=0;j<=m;j++)
 76 //           printf("%d ",A[i][j]);
 77 //       printf("\n");
 78 //    }
 79     int k,col,num=0,max_r,dou,max_x,LCM,ta,tb;
 80     for(k=0,col=0;k<n && col<m;k++,col++)
 81     {
 82         max_r=k;
 83         max_x=abs(A[k][col]);
 84         for(int i=k+1;i<n;i++)
 85             if(max_x<abs(A[i][col]))
 86             {
 87                 max_x=abs(A[i][col]);
 88                 max_r=i;
 89             }
 90         if(max_r!=k)
 91         {
 92             for(int j=col;j<=m;j++)
 93                 swap(A[k][j],A[max_r][j]);
 94         }
 95         if(A[k][col]==0)
 96         {
 97             k--;
 98             free_x[num++]=col;
 99             continue;
100         }
101         for(int i=k+1;i<n;i++)
102         if(A[i][col])
103         {
104             LCM=lcm(A[k][col],A[i][col]);
105             ta=LCM/A[i][col];
106             tb=LCM/A[k][col];
107             for(int j=col;j<=m;j++)
108             {
109                     A[i][j]=((A[i][j]*ta-A[k][j]*tb)%3+3)%3;
110             }
111         }
112     }
113     int temp;
114     for(int i=0;i<num;i++)
115         x[free_x[i]]=0;
116     int  xi,yi;
117     for(int i=k-1,c=m-1;i>=0;c=m-1,i--)
118     {
119         temp=A[i][m];
120         while(x[c]!=-1)
121         {
122             if(A[i][c])
123                 temp=((temp-(x[c]*A[i][c])%3)%3+3)%3;
124             c--;
125         }
126         exgcd(A[i][c],3,xi,yi);
127         xi=(xi%3+3)%3;
128         x[c]=(temp*xi%3+3)%3;
129     }
130 //    for(int i=0;i<n;i++)
131 //    {
132 //        for(int j=0;j<=m;j++)
133 //            printf("%d ",A[i][j]);
134 //        printf("\n");
135 //    }
136 //    for(int i=0;i<m;i++)
137 //        printf("%d ",x[i]);
138     return ;
139 }
140 int exgcd(int a,int b,int &x,int &y)
141 {
142     if(b==0)
143     {
144         x=1;
145         y=0;
146         return a;
147     }
148     else
149     {
150         int r=exgcd(b,a%b,y,x);
151         y-=x*(a/b);
152         return r;
153     }
154 }
155 int gcd(int a,int b)
156 {
157     int c;
158     while(b!=0)
159     {
160         c=a%b;
161         a=b;
162         b=c;
163     }
164     return a;
165 }
166 int lcm(int a,int b)
167 {
168     return a/gcd(a,b)*b;
169 }

 

hdu 5755(GAuss 消元)

标签:

原文地址:http://www.cnblogs.com/wujiechao/p/5862935.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!