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

JSZKC is going to spend his vacation!

His vacation has

To avoid this problem, he has

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
The next follows $j^{th}jth integer in the i^{th}ith line means [i][j] 1≤f[i][j]≤1000000).$

There are no more than

Output Format

One line per case, an integer indicates the answer.

样例输入

样例输出

2
9

题目来源

The 2018 ACM-ICPC China JiangSu Provincial Programming Contest

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <queue>
 7 #include <set>
 8 #include <map>
 9 #include <string>
10 #include <cmath>
11 #include <cstdlib>
12 #include <ctime>
13 using namespace std;
14 typedef long long ll;
15 /*
16 f[a][b]+f[b][c]+f[c][d]+……+f[y][z]的最大值。(n-1项)
17 n=2 :二重循环
18 n=3 :三重循环
19 n=4 : f[a][b]+f[b][c]+f[c][d]
20 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]
21 此时的f[a][c]表示第一天a,第三天c的最大值
22 n>=5的依此类推
23 那么可以利用矩阵快速幂的思想，因为(n-2)*m^2会超时
24 n=2要更新一次来找最大值,n=3就要更新2次。
25 因此n要更新n-1次。
26 */
27 const int  N=110;
28 ll  f[N][N],n,m;
29 struct ma{
30     ll m[N][N];
31      ma(){
32          memset(m,0,sizeof(m));
33      }
34 };
35 ll MAX;
36 ma poww(ma a,ma  b)
37 {
38     ma c;
39     for(int i=0;i<m;i++)
40     {
41         for(int j=0;j<m;j++)
42            {
43         for(int k=0;k<m;k++)
44         {
45             c.m[i][j]=max(c.m[i][j],a.m[i][k]+b.m[k][j]);
46         }
47            }
48     }
49     return c;
50 }
51 ma qu(ma a,ll n){
52     ma c;
53     while(n){
54         if(n&1) c=poww(c,a);
55           n>>=1;
56           a=poww(a,a);
57     }
58     return c;
59 }
60 int main()
61 {
62     while(~scanf("%lld%lld",&n,&m)){
63           ma ans;
64         for(int i=0;i<m;i++)
65         {
66             for(int j=0;j<m;j++)
67             {
68                 scanf("%lld",&ans.m[i][j]);
69             }
70         }
71         ans=qu(ans,n-1);
72         MAX=0;        
73         for(int i=0;i<m;i++){
74             for(int j=0;j<m;j++)
75             {
76                 MAX=max(MAX,ans.m[i][j]);
77             }
78         }
79         printf("%lld\n",MAX);
80     }
81     return 0;
82 }