[矩阵快速幂]T-shirt(2018江苏邀请赛I题)

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

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

题意：求f[a][b]+f[b][c]+f[c][d]+...+f[p][q](有n-1项)的最大值；
思路：从简单情况入手，当n=2时，求f[a][b]的最大值，两重循环遍历即可；
当n=3时，求f[a][b]+f[b][c]的最大值，三重循环遍历即可；
当n=4时，自然可以四重循环遍历，，，我们换一个角度考虑，可以将f[a][c]的值以f[a][b]+f[b][c]中的最大值替代(得到新的最优的f[1~m][1~m]——此时f[i][j]表示第一天穿i,第三天穿j可以得到的最大快乐)，接着求f[a][c]+f[c][d]的最大值，回到n=3的情况，也就是进行三重循环遍历；
当n=5时类似，将f[a][d]以f[a][c]+f[c][d]中的最值替代后，计算f[a][d]+f[d][e]的最值，与此同时，得到新的f[1~m][1~m];
总结一下，相当于不断地更新f[1~m][1~m]；一开始f[i][j]表示第1天穿i,第2天穿j的所能得到的最大快乐，更新k次后f[i][j]表示第1天穿i，第2+k天穿j所能得到的最大快乐；
这样的话需要在原始f上更新n-2次得到最终f——f[i][j]表示第一天穿i，第n天穿j所能得到的最大快乐；每一次更新的复杂度为O(m^2)，暴力n次更新复杂度O(n*m^2)会超时；
可以将更新视为定义在f上的一种运算‘#‘，f经n-2次更新，可视为f#f#...#f(n-1个f相#)，用“快速#”计算n-1个f相#即可；类似于矩阵快速幂；

AC代码：

#include <iostream> #include<cstdio> #include<cstring> typedef long long ll; using namespace std; ll n,m; struct Matrix{ ll f[110][110]; Matrix(){memset(f,0,sizeof(f));} Matrix operator *(Matrix & mat){ Matrix ret; for(ll i=1;i<=m;i++){ for(ll k=1;k<=m;k++){ for(ll j=1;j<=m;j++){ ret.f[i][j]=max(ret.f[i][j],mat.f[i][k]+f[k][j]); } } } return ret; } }; Matrix qpow(Matrix a,ll b){ Matrix ret; while(b){ if(b&1) ret=ret*a; a=a*a; b>>=1; } return ret; } int main() { while(scanf("%lld%lld",&n,&m)!=EOF){ Matrix mat; for(ll i=1;i<=m;i++){ for(ll j=1;j<=m;j++){ scanf("%lld",&mat.f[i][j]); } } mat=qpow(mat,n-1); ll ans=0; for(ll i=1;i<=m;i++){ for(ll j=1;j<=m;j++){ ans=max(ans,mat.f[i][j]); } } printf("%lld\n",ans); } return 0; }