徐州邀请赛江苏 icpc I. T-shirt 矩阵快速幂-优快云博客

本文解析了一道关于寻找最大权值路径的算法题，通过建立动态规划模型并运用矩阵快速幂的方法来求解最优解。适用于解决具有周期性质的最优化问题。

题目

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

样例输出

2
9

题目来源

The 2018 ACM-ICPC China JiangSu Provincial Programming Contest

题意：有n天，m件衣服，如果某一天穿了第i件衣服，第二天穿了第j件衣服，那么就会获得dp[i][j]的权值。然后给你矩阵f，每件衣服可以穿无限多次，问第n天能获得的最大权值是多少。

分析：设

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 107;
const int mod = 1e9 + 7;
typedef long long ll;
struct matrix {
    ll a[maxn][maxn];
};
ll n, m;
matrix base, ans;
matrix mul( matrix x, matrix y ) {
    matrix tmp;
    memset( tmp.a, 0, sizeof(tmp.a) );
    for( ll i = 0; i < m; i ++ ) {
        for( ll j = 0; j < m; j ++ ) {
            for( ll k = 0; k < m; k ++ ) {
                tmp.a[i][j] = max( tmp.a[i][j], x.a[i][k] + y.a[k][j] );
            }
        }
    }
    return tmp;
}
void qow( ll x ) {
    while( x ) {
        if( x&1 ) {
            ans = mul( ans, base );
        }
        base = mul( base, base );
        x /= 2;
    }
}
int main() {
    std::ios::sync_with_stdio(false);
    while( cin >> n >> m ) {
        memset( ans.a, 0, sizeof(ans.a) );
        for( ll i = 0; i < m; i ++ ) {
            for( ll j = 0; j < m; j ++ ) {
                cin >> base.a[i][j];
            }
        }
        qow( n-1 );
        ll sum = 0;
        for( ll i = 0; i < m; i ++ ) {
            for( ll j = 0; j < m; j ++ ) {
                //cout << ans.a[i][j] << " ";
                sum = max( sum, ans.a[i][j] );
            }
           // cout << endl;
        }
        cout << sum << endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/l609929321/p/9364641.html