T-shirt

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

感觉蛮难的一道题，学长给我讲了还迷迷糊糊的，看了半小时才明白。
这题用的东西蛮多的，首先用了倍增的思想，f[i][j][k]表示走2^i次，从j->k的最大价值。也就是压缩了一下状态。
然后用了二进制的思想，把n分解成x1*2^p1+x2*2^p2+x3*2^p3.....xn*2^pn。 然后看能不能xi是否等于0，如果等于，可以走。 就进入递推，这样保证了一定走了n-1次。
然后用了一个状态压缩，因为每一步只和上一步有关，所以i只需要在0与1之间变化。

#include <bits/stdc++.h>
#define maxn 105
using namespace std;
typedef long long ll;
ll f[25][maxn][maxn]={0};
ll ans[2][maxn][maxn]={0};
int main()
{
    int n,m,i,j,k,l;
    while(~scanf("%d%d",&n,&m))
    {
        memset(f,0,sizeof(f));
        memset(ans,0,sizeof(ans));
        for(i=1;i<=m;i++)
        {
            for(j=1;j<=m;j++)
            {
                scanf("%lld",&f[0][i][j]);
            }
        }
        for(i=1;i<=20;i++)
        {
            for(j=1;j<=m;j++)
            {
                for(k=1;k<=m;k++)
                {
                    for(l=1;l<=m;l++)
                    {
                        f[i][j][k]=max(f[i][j][k],f[i-1][j][l]+f[i-1][l][k]);
                    }
                }
            }
        }
        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]+f[i][l][k]);
                    }
                }
            }
            temp=1-temp;
            }
        }
        temp=1-temp;
        ll maxim=0;
        for(i=1;i<=m;i++)
        {
            for(j=1;j<=m;j++)
            {
                maxim=max(ans[temp][i][j],maxim);
            }
        }
        cout<<maxim<<endl;
    }
    return 0;
}