The 2018 ACM-ICPC China JiangSu Provincial Programming Contest I. T-shirt
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 Bcolor 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.
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,M (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 [i][j] 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
3 2
0 1
1 0
4 3
1 2 3
1 2 3
1 2 3
样例输出
2
9
题目来源
The 2018 ACM-ICPC China JiangSu Provincial Programming Contest
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
typedef long long ll;
/*
f[a][b]+f[b][c]+f[c][d]+……+f[y][z]的最大值。(n-1项)
n=2 :二重循环
n=3 :三重循环
n=4 : f[a][b]+f[b][c]+f[c][d]
可以用f[a][c]来代替f[a][c]和f[a][b]+f[b][c]的较大值,在进行f[a][c]+f[c][d]
此时的f[a][c]表示第一天a,第三天c的最大值
n>=5的依此类推
那么可以利用矩阵快速幂的思想,因为(n-2)*m^2会超时
n=2要更新一次来找最大值,n=3就要更新2次。
因此n要更新n-1次。
*/
const int N=;
ll f[N][N],n,m;
struct ma{
ll m[N][N];
ma(){
memset(m,,sizeof(m));
}
};
ll MAX;
ma poww(ma a,ma b)
{
ma c;
for(int i=;i<m;i++)
{
for(int j=;j<m;j++)
{
for(int k=;k<m;k++)
{
c.m[i][j]=max(c.m[i][j],a.m[i][k]+b.m[k][j]);
}
}
}
return c;
}
ma qu(ma a,ll n){
ma c;
while(n){
if(n&) c=poww(c,a);
n>>=;
a=poww(a,a);
}
return c;
}
int main()
{
while(~scanf("%lld%lld",&n,&m)){
ma ans;
for(int i=;i<m;i++)
{
for(int j=;j<m;j++)
{
scanf("%lld",&ans.m[i][j]);
}
}
ans=qu(ans,n-);
MAX=;
for(int i=;i<m;i++){
for(int j=;j<m;j++)
{
MAX=max(MAX,ans.m[i][j]);
}
}
printf("%lld\n",MAX);
}
return ;
}
The 2018 ACM-ICPC China JiangSu Provincial Programming Contest I. T-shirt的更多相关文章
- The 2018 ACM-ICPC China JiangSu Provincial Programming Contest快速幂取模及求逆元
题目来源 The 2018 ACM-ICPC China JiangSu Provincial Programming Contest 35.4% 1000ms 65536K Persona5 Per ...
- The 2018 ACM-ICPC China JiangSu Provincial Programming Contest J. Set
Let's consider some math problems. JSZKC has a set A=A={1,2,...,N}. He defines a subset of A as 'Meo ...
- The 2018 ACM-ICPC China JiangSu Provincial Programming Contest(第六场)
A Plague Inc Plague Inc. is a famous game, which player develop virus to ruin the world. JSZKC wants ...
- C.0689-The 2019 ICPC China Shaanxi Provincial Programming Contest
We call a string as a 0689-string if this string only consists of digits '0', '6', '8' and '9'. Give ...
- B.Grid with Arrows-The 2019 ICPC China Shaanxi Provincial Programming Contest
BaoBao has just found a grid with $n$ rows and $m$ columns in his left pocket, where the cell in the ...
- ACM ICPC, Damascus University Collegiate Programming Contest(2018) Solution
A:Martadella Stikes Again 水. #include <bits/stdc++.h> using namespace std; #define ll long lon ...
- 计蒜客 39272.Tree-树链剖分(点权)+带修改区间异或和 (The 2019 ACM-ICPC China Shannxi Provincial Programming Contest E.) 2019ICPC西安邀请赛现场赛重现赛
Tree Ming and Hong are playing a simple game called nim game. They have nn piles of stones numbered ...
- 计蒜客 39280.Travel-二分+最短路dijkstra-二分过程中保存结果,因为二分完最后的不一定是结果 (The 2019 ACM-ICPC China Shannxi Provincial Programming Contest M.) 2019ICPC西安邀请赛现场赛重现赛
Travel There are nn planets in the MOT galaxy, and each planet has a unique number from 1 \sim n1∼n. ...
- 计蒜客 39279.Swap-打表找规律 (The 2019 ACM-ICPC China Shannxi Provincial Programming Contest L.) 2019ICPC西安邀请赛现场赛重现赛
Swap There is a sequence of numbers of length nn, and each number in the sequence is different. Ther ...
随机推荐
- shell 发送Post请求,并获取状态码
#!/bin/bash aa=$ result=$(curl -H "Content-type: application/json" -X POST -o /dev/null -s ...
- (转)nginx域名访问的白名单配置梳理
nginx域名访问的白名单配置梳理 原文:http://www.cnblogs.com/kevingrace/p/6086652.html 在日常运维工作中,会碰到这样的需求:设置网站访问只对某些ip ...
- (转)Centos 7.3 用户和组管理
Centos 7.3 用户和组管理 原文:http://blog.csdn.net/github_39069288/article/details/73306489 3.1 用户和密码配置文件 pas ...
- 对于es线程池使用的思考
es有内置的线程池 在实际项目中,发现 使用client框架关闭连接太慢(其实是把连接归还到池子里),采用异步关闭. 随着连接的关闭,计算机内存在不断下降 ------------------- ...
- Elasticsearch优化
2.out of memory错误 因为默认情况下es对字段数据缓存(Field Data Cache)大小是无限制的,查询时会把字段值放到内存,特别是facet查询,对内存要求非常高,它会把结果都放 ...
- php分页代码及总结
代码部分: <?PHPheader("Content-type:text/html;charset=utf-8");$pageSize = 10;//接收传入的分页码$pag ...
- 关于如何将html中的表格下载成csv格式的方法
今天在网上看了很多方法,自己还是慢慢探索写出了最终效果 简单代码如下: <!DOCTYPE html> <html> <head> <meta content ...
- 如何设置FusionCharts图片导出格式
通过设置FusionCharts的<chart exportEnabled='1' ...>属性,就可以导出图表,图表的右键菜单将会显示所有可能导出的格式- JPEG, PNG and P ...
- Burpsuite Professional安装及使用教程
转自:https://www.jianshu.com/p/edbd68d7c341 1.先从吾爱破解论坛下载工具:https://down.52pojie.cn/Tools/Network_Analy ...
- 用python Image读图
https://www.cnblogs.com/kongzhagen/p/6295925.html import os name = [] with open('/media/hdc/xing/Dee ...