T-shirt

时间限制: 1 Sec 内存限制: 64 MB

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

直接套矩阵快速幂的板子就行

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

typedef long long ll;
using namespace std;
const ll maxn = 110;
const ll mod = 1e9+7;

struct Mar{
    ll f[maxn][maxn];
    ll m;
    Mar(ll mm){
        m = mm;
        memset(f,0,sizeof f);
    }
    Mar operator*(const  Mar& b){
        Mar ret(m);
        for (int i=1;i<=m;i++){
            for (int j=1;j<=m;j++){
                for (int k=1;k<=m;k++){
                    ret.f[i][j]=max(ret.f[i][j],f[i][k]+b.f[k][j]);
                }
            }
        }
        return ret;
    }
    Mar pow(ll x){
        Mar ret(m);
        Mar a = *this;
        while(x){
            if (x&1ll)
                ret = ret*a;
            a=a*a;
            x>>=1;
        }
        return ret;
    }
};

int main(){
    std::ios::sync_with_stdio(false);
    std::cin.tie();
    ll k,m;
    while(cin>>k>>m){
        Mar f(m);
        for (int i=1;i<=m;i++)
            for (int j=1;j<=m;j++)
                cin>>f.f[i][j];
        f = f.pow(k-1);
        ll ans = 0;
        for (int i=1;i<=m;i++)
            for (int j=1;j<=m;j++)
                ans = max(ans,f.f[i][j]);
        cout<return 0;
}

/**************************************************************
    Problem: 7561
    User: WC011
    Language: C++
    Result: 正确
    Time:416 ms
    Memory:2036 kb
****************************************************************/