The 2018 ACM-ICPC China JiangSu Provincial Programming Contest I. T-shirt

JSZKC is going to spend his vacation!

His vacation has N days. Each day, he can choose a T-shirt to wear. Obviously, he doesn't want to wear a singer color T-shirt since others will consider he has worn one T-shirt all the time.

To avoid this problem, he has M different T-shirt with different color. If he wears A color T- shirt this day and Bcolor T-shirt the next day, then he will get the pleasure of f[[A][B].(notice: He is able to wear one T-shirt in two continuous days but may get a low pleasure)

Please calculate the max pleasure he can get.

Input Format

The input file contains several test cases, each of them as described below.

• The first line of the input contains two integers N,M (2≤N≤100000,1≤M≤100), giving the length of vacation and the T-shirts that JSZKC has.

• The next follows MM lines with each line MM integers. The j^{th}jth integer in the i^{th}ith line means [i][j] 1≤f[i][j]≤1000000).

There are no more than 1010 test cases.

Output Format

One line per case, an integer indicates the answer.

样例输入

3 2
0 1
1 0
4 3
1 2 3
1 2 3
1 2 3

样例输出

2
9

题目来源

The 2018 ACM-ICPC China JiangSu Provincial Programming Contest

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <cmath>
 #include <cstdlib>
 #include <ctime>
 using namespace std;
 typedef long long ll;
 /*
 f[a][b]+f[b][c]+f[c][d]+……+f[y][z]的最大值。(n-1项)
 n=2 :二重循环
 n=3 :三重循环
 n=4 : f[a][b]+f[b][c]+f[c][d]
 可以用f[a][c]来代替f[a][c]和f[a][b]+f[b][c]的较大值，在进行f[a][c]+f[c][d]
 此时的f[a][c]表示第一天a,第三天c的最大值
 n>=5的依此类推
 那么可以利用矩阵快速幂的思想，因为(n-2)*m^2会超时
 n=2要更新一次来找最大值,n=3就要更新2次。
 因此n要更新n-1次。
 */
 const int  N=;
 ll  f[N][N],n,m;
 struct ma{
     ll m[N][N];
      ma(){
          memset(m,,sizeof(m));
      }
 };
 ll MAX;
 ma poww(ma a,ma  b)
 {
     ma c;
     for(int i=;i<m;i++)
     {
         for(int j=;j<m;j++)
            {
         for(int k=;k<m;k++)
         {
             c.m[i][j]=max(c.m[i][j],a.m[i][k]+b.m[k][j]);
         }
            }
     }
     return c;
 }
 ma qu(ma a,ll n){
     ma c;
     while(n){
         if(n&) c=poww(c,a);
           n>>=;
           a=poww(a,a);
     }
     return c;
 }
 int main()
 {
     while(~scanf("%lld%lld",&n,&m)){
           ma ans;
         for(int i=;i<m;i++)
         {
             for(int j=;j<m;j++)
             {
                 scanf("%lld",&ans.m[i][j]);
             }
         }
         ans=qu(ans,n-);
         MAX=;
         for(int i=;i<m;i++){
             for(int j=;j<m;j++)
             {
                 MAX=max(MAX,ans.m[i][j]);
             }
         }
         printf("%lld\n",MAX);
     }
     return ;
 }