题目描述

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 B color 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. 

输入

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 M lines with each line M integers. The jth integer in the ith line means f[i][j](1<=f[i][j]<=1000000). 
There are no more than 10 test cases. 

输出

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
题解
看了大佬的代码，迷迷糊糊的
题意很好理解 
这里用了倍增的思想
倍增思想之前接触过一道题
f[i][j][k]表示走2^i次，从j->k的最大价值，状态压缩？
然后用了二进制按位与
就是二进制位上都为1的话就说明走了这个，可以递推
n分解成x1*2^p1+x2*2^p2+x3*2^p3.....xn*2^pn
就看xi是不是为1
最后 每一步只与上一步有关 就可以0和1表示
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=105;
ll s[25][maxn][maxn];//s[a][b][c] b->c 2^a days
ll ans[2][maxn][maxn];
int m,n;
int main(){
    int i,j,l,k;
    while(cin>>n>>m){
    memset(s,0,sizeof(s));
    memset(ans,0,sizeof(ans));
    for(i=1;i<=m;i++){
        for(j=1;j<=m;j++){
            cin>>s[0][i][j];
        }
    }
    for(i=1;i<=20;i++){
        for(j=1;j<=m;j++){
            for(l=1;l<=m;l++){
                for(k=1;k<=m;k++){
                    s[i][j][l]=max(s[i][j][l],s[i-1][j][k]+s[i-1][k][l]);
                }
            }
        }
    }
    n--;
    int temp=1;
    for(i=0;i<20;i++){
        if(n&(1<<i)){
             for(j=1;j<=m;j++){
                for(k=1;k<=m;k++){
                    for(l=1;l<=m;l++){
                        ans[temp][j][k]=max(ans[temp][j][k],ans[1-temp][j][l]+s[i][l][k]);
                    }
                }
            }
            temp=1-temp;
        }
    }
    temp=1-temp;
    ll maxx=0;
     for(i=1;i<=m;i++){
            for(j=1;j<=m;j++){
                maxx=max(ans[temp][i][j],maxx);
            }
        }
        cout<<maxx<<endl;
    }
    return 0;
}
View Code

这个是状态压缩DP：https://www.cnblogs.com/ibilllee/p/7651971.html