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=110;
ll  f[N][N],n,m;
struct ma{
    ll m[N][N];
     ma(){
         memset(m,0,sizeof(m));
     }
};
ll MAX;
ma poww(ma a,ma  b)
{
    ma c;
    for(int i=0;i<m;i++)
    {
        for(int j=0;j<m;j++)
           {
        for(int k=0;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&1) c=poww(c,a);
          n>>=1;
          a=poww(a,a);
    }
    return c;
}
int main()
{
    while(~scanf("%lld%lld",&n,&m)){
          ma ans;
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<m;j++)
            {
                scanf("%lld",&ans.m[i][j]);
            }
        }
        ans=qu(ans,n-1);
        MAX=0;        
        for(int i=0;i<m;i++){
            for(int j=0;j<m;j++)
            {
                MAX=max(MAX,ans.m[i][j]);
            }
        }
        printf("%lld
",MAX);
    }
    return 0;
}