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Codeforces Round #264 (Div. 2) C Gargari and Bishops 【暴力】
2022-07-06 19:37:00 【全栈程序员站长】
大家好,又见面了,我是全栈君
称号:
意甲冠军:给定一个矩阵,每格我们有一个数,然后把两个大象,我希望能够吃的对角线上的所有数字。我问两个最大的大象可以吃值。
分析:这种想法是暴力的主题,计算出每一格放象的话能得到多少钱,然后求出两个不冲突的最大值。我比赛的时候写的方法是先求出每个值,编号之后存到数组里面,然后在通过一系列处理得到,总之非常麻烦。写了一个多小时。最后才发现有一点小错误,没时间了。初始例子也没有通过。
然后看了下别人写的,太简洁了。直接用行和列和和差就能够直接求出来。看来代码能力还很有待提高啊。
我的AC代码:
#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;
}
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