2018徐州邀请赛I题 I. T-shirt

JSZKC is going to spend his vacation!

His vacation has NN 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 MM different T-shirt with different color. If he wears AA color T- shirt this day and BBcolor T-shirt the next day, then he will get the pleasure of f[A][B]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.

Input Format

The input file contains several test cases, each of them as described below.

The first line of the input contains two integers N,MN,M (2 \le N \le 100000, 1 \le M \le 100)(2≤N≤100000,1≤M≤100), giving the length of vacation and the T-shirts that JSZKC has.
The next follows MM lines with each line MM integers. The j^{th}jth integer in the i^{th}ith line means f[i][j]f[i][j] (1\le f[i][j]\le 1000000)(1≤f[i][j]≤1000000).

There are no more than 1010 test cases.

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

思路：本题就是让你求一个n-1的序列。f[a][b]+f[b][c]+f[c][d]+~~~~+ f[x][y]+f[y][z]的最大值。

那么第0 天（也就是题目中的第一天）可以穿任意的衣服，那么第一天就根据第0 天的衣服确定最大价值，我们可以设dp[time][i][j] 表示第0天穿i 第time天穿j 的最大价值，那么我们就可以得出 dp[a+b][i][j] =max(dp[a][i][k]+dp[b][k][j])。那么对于该式子我们可以用快速幂处理来降低时间复杂度。

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
const int N =105;

struct node
{
    ll ma[N][N];
};
int n,m;

node mut(node a,node b)
{
    node ans;
    memset(ans.ma,0,sizeof(ans.ma));
    for(int i=1;i<=m;i++){
        for(int j=1;j<=m;j++){
            for(int k=1;k<=m;k++){
                ans.ma[i][j]=max(ans.ma[i][j],a.ma[i][k]+b.ma[k][j]);
            }
        }
    }
    return ans;
}

node quick_pow(node a,int k)
{
    node ans;
    memset(ans.ma,0,sizeof(ans.ma));
    while(k)
    {
        if(k&1) ans=mut(ans,a);
        a=mut(a,a);
        k>>=1;
    }
    return ans;
}

int main()
{
    node tmp;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        for(int i=1;i<=m;i++){
            for(int j=1;j<=m;j++){
                scanf("%lld",&tmp.ma[i][j]);
            }
        }

        node ans=quick_pow(tmp,n-1);

        ll maxx=0;
        for(int i=1;i<=m;i++){
            for(int j=1;j<=m;j++){
                maxx=max(maxx,ans.ma[i][j]);
            }
        }
        printf("%lld\n",maxx);
    }

    return 0;
}