当前位置:网站首页>Codeforces round 264 (Div. 2) C gargari and Bishop [violence]
Codeforces round 264 (Div. 2) C gargari and Bishop [violence]
2022-07-07 03:10:00 【Full stack programmer webmaster】
Hello everyone , I meet you again , I'm the king of the whole stack
The title of :
Serie A Champion : Given a matrix , We have a number in each grid , Then put two elephants , I hope to be able to eat all the numbers on the diagonal . I asked the two biggest elephants what they could eat .
analysis : This idea is the subject of violence , Calculate how much money you can get for each grid , Then find two non conflicting maxima . The method I wrote during the competition is to find out each value first , Number and save it in the array , Then through a series of processing , In short, it's very troublesome . For more than an hour . Finally, I found a little mistake , There's no time . The initial example also failed .
Then I read what others wrote , It's too simple . You can find it directly by using the sum and difference of rows and columns . It seems that the code ability still needs to be improved .
my AC Code :
#include <cstdio>
#include <iostream>
#include <queue>
#include <cstring>
#include <algorithm>
#include <stack>
#include <vector>
#include <utility>
#include <cmath>
using namespace std;
const long long N = 2200;
long long mp[N][N],num[N][N];
vector<long long> sum1,sum2;
long long n;
int main()
{
//freopen("Input.txt","r",stdin);
while(~scanf("%lld",&n))
{
for(long long i=1; i<=n; i++)
{
for(long long j=1; j<=n; j++)
scanf("%lld",&mp[i][j]);
}
for(long long i=n; i>=1; i--)
{
long long tmp=0;
for(long long x=1,y=i; x<=n; x++,y++)
{
tmp+=mp[x][y];
}
sum1.push_back(tmp);
}
for(long long i=2; i<=n; i++)
{
long long tmp=0;
for(long long x=i,y=1; x<=n; x++,y++)
tmp+=mp[x][y];
sum1.push_back(tmp);
}
for(long long i=1; i<=n; i++)
{
long long tmp=0;
for(long long x=i,y=1; x>=1; x--,y++)
{
tmp+=mp[x][y];
}
sum2.push_back(tmp);
}
for(long long i=2; i<=n; i++)
{
long long tmp=0;
for(long long x=n,y=i; y<=n; x--,y++)
tmp+=mp[x][y];
sum2.push_back(tmp);
}
memset(num,0,sizeof(num));
long long fx=1,fy=n;
for(long long i=1;i<=n;i++)
{
long long xx=fx,yy=fy;
for(long long j=1;j<=n;j++)
{
num[i][j]=sum2[xx-1]+sum1[yy-1]-mp[i][j];
xx++,yy--;
//printf("%lld ",num[i][j]);
}
fx++,fy++;
}
long long ans[3]={-1,-1};
long long x[3],y[3];
for(long long i=1;i<=n;i++)
{
for(long long j=1;j<=n;j++)
{
long long p=(i+j)%2;
if(num[i][j]>ans[p])
{
ans[p]=num[i][j];
x[p]=i,y[p]=j;
}
}
}
printf("%lld\n",ans[0]+ans[1]);
printf("%lld %lld %lld %lld\n",x[0],y[0],x[1],y[1]);
sum1.clear(),sum2.clear();
}
return 0;
}
Copyright notice : This article is the original article of the blogger , Blog , Do not reprint without permission .
Publisher : Full stack programmer stack length , Reprint please indicate the source :https://javaforall.cn/116787.html Link to the original text :https://javaforall.cn
边栏推荐
- Contribution of Writing Series
- Oauth2协议中如何对accessToken进行校验
- INS/GPS组合导航类型简介
- Redis入门完整教程:RDB持久化
- 【Socket】①Socket技术概述
- Laravel php artisan 自动生成Model+Migrate+Controller 命令大全
- SQL Tuning Advisor一个错误ORA-00600: internal error code, arguments: [kesqsMakeBindValue:obj]
- Lingyun going to sea | yidiantianxia & Huawei cloud: promoting the globalization of Chinese e-commerce enterprise brands
- Shell 编程基础
- How-PIL-to-Tensor
猜你喜欢
MOS transistor realizes the automatic switching circuit of main and auxiliary power supply, with "zero" voltage drop and static current of 20ua
QT Bluetooth: qbluetooth DeviceInfo
Nuggets quantification: obtain data through the history method, and use the same proportional compound weight factor as Sina Finance and snowball. Different from flush
[secretly kill little partner pytorch20 days] - [Day1] - [example of structured data modeling process]
掘金量化:通过history方法获取数据,和新浪财经,雪球同用等比复权因子。不同于同花顺
uniapp的表单验证
Google Earth engine (GEE) -- 1975 dataset of Landsat global land survey
杰理之开启经典蓝牙 HID 手机的显示图标为键盘设置【篇】
ERROR: Could not find a version that satisfies the requirement xxxxx (from versions: none)解决办法
Es6中Promise的使用
随机推荐
Lingyun going to sea | yidiantianxia & Huawei cloud: promoting the globalization of Chinese e-commerce enterprise brands
Detailed explanation of 19 dimensional integrated navigation module sinsgps in psins (time synchronization part)
Analysis of USB network card sending and receiving data
Redis入门完整教程:复制拓扑
Redis Getting started tutoriel complet: positionnement et optimisation des problèmes
PSINS中19维组合导航模块sinsgps详解(时间同步部分)
从控制理论的角度谈数据分析
杰理之关于 DAC 输出功率问题【篇】
HDU 4337 King Arthur&#39;s Knights 它输出一个哈密顿电路
Redis getting started complete tutorial: common exceptions on the client
New benchmark! Intelligent social governance
MOS transistor realizes the automatic switching circuit of main and auxiliary power supply, with "zero" voltage drop and static current of 20ua
Oracle connection pool is not used for a long time, and the connection fails
Cryptography series: detailed explanation of online certificate status protocol OCSP
The version control of 2021 version is missing. Handling method
Redis getting started complete tutorial: replication configuration
Examples of how to use dates in Oracle
Laravel php artisan 自动生成Model+Migrate+Controller 命令大全
Redis introduction complete tutorial: client case analysis
【Socket】①Socket技术概述