问题 I: T-shirt

问题 I: T-shirt

时间限制: 1 Sec  内存限制: 64 MB
提交: 14  解决: 7
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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

提示

思路： 我的思路类似于最短路的floyd ， 对每天的每件物品求一下对m件衣服的权值和，保留一个最大值。

因为只需要选n-1天，所以用一个快速幂就可以了。

代码如下：

#include <bits/stdc++.h>
using namespace std;
const int maxn=110;
typedef long long ll;
ll n,m;
struct node{
    ll f[maxn][maxn];
    node(){
        memset(f,0,sizeof(f));
    }
    node operator*(const node& p){
        node ans;
        for (int i=1; i<=m ;i++)
            for (int j=1; j<=m ;j++)
                for (int k=1; k<=m ;k++)
                    ans.f[i][j]=max(ans.f[i][j],f[i][k]+p.f[k][j]);
        return ans;
    }
    node Pow_mod(ll n){
        node ans;
        node p=(*this);
        while(n){
            if(n&1) ans=ans*p;
            p=p*p;
            n>>=1;
        }
        return ans;
    }
};

int main(){
    while(scanf("%lld%lld",&n,&m)==2){
        node mat;
        for (int i=1; i<=m ;i++)
            for (int j=1; j<=m; j++)
                scanf("%lld",&mat.f[i][j]);
        mat=mat.Pow_mod(n-1);
        ll ans=0;
        for (int i=1; i<=m ;i++)
            for (int j=1; j<=m ;j++)
                ans=max(ans,mat.f[i][j]);
        printf("%lld
",ans);
    }
    return 0;
}