【倍增】T-shirt @2018acm徐州邀请赛 I

问题 I: 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

样例输出

2
9

meaning

给一个m×m的矩阵，data[i][j]表示第一天在i点第二天在j点的收益，问n天的最大收益。

solution

f[k][i][j] 表示第一天在i点，2^k天后在j点的最大收益。

f[0][i][j] = data[i][j];

f[k][i][j] = max(f[k][i][j],f[k-1][i][p]+f[k-1][p][j]);

d[i][j] 表示从i点到j点的最大收益。

code

#define IN_LB() freopen("C:\Users\acm2018\Desktop\in.txt","r",stdin)
#define OUT_LB() freopen("C:\Users\acm2018\Desktop\out.txt","w",stdout)
#define IN_PC() freopen("C:\Users\hz\Desktop\in.txt","r",stdin)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxm = 105;

ll f[20][maxm][maxm];
ll d[2][maxm][maxm];
int main() {
//    IN_LB();
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF) {
        memset(f,0,sizeof f);
        memset(d,0,sizeof d);
        for(int i=0; i<m; i++) {
            for(int j=0; j<m; j++) {
                scanf("%lld",&f[0][i][j]);
            }
        }
        for(int k=1; k<20; k++)
            for(int l=0; l<m; l++)
                for(int i=0; i<m; i++)
                    for(int j=0; j<m; j++)
                        f[k][i][j] = max(f[k][i][j],f[k-1][i][l]+f[k-1][l][j]);
        n--;
        int cnt = 0;
        for(int k=19; k>=0; k--) {
            if(n>(1<<k)) {
                n-=(1<<k);
                for(int l = 0; l<m; l++) {
                    for(int i=0; i<m; i++) {
                        for(int j=0; j<m; j++) {
                            d[cnt][i][j]=max(d[cnt][i][j],d[cnt^1][i][l]+f[k][l][j]);
                        }
                    }
                }
                cnt^=1;
            }
        }
        if(n==1) {
            for(int l = 0; l<m; l++) {
                for(int i=0; i<m; i++) {
                    for(int j=0; j<m; j++) {
                        d[cnt][i][j]=max(d[cnt][i][j],d[cnt^1][i][l]+f[0][l][j]);
                    }
                }
            }
        }
        ll ans = 0;
        for(int i=0; i<m; i++) {
            for(int j=0; j<m; j++) {
                ans = max(ans,d[cnt][i][j]);
            }
        }
        printf("%lld
",ans);
    }
    return 0;
}