版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/violinlove/article/details/81673588
emmmm,不知道说啥,但是还是改了很久,好想,不过看来容斥运用地还不到位,真是浪费 LZH 小朋友 的一番苦心啊
还有 边界问题,细节问题 以及 不成熟的想法(容斥不能乱用)!!
附上代码:
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
inline int wread(){
char c(getchar ());int wans (0),flag(1);
while (c<'0' ||c>'9'){if (c=='-') flag=-1; c=getchar ();}
while (c>='0' && c<='9'){wans=wans*10+c-'0';c=getchar ();}
return wans*=flag;
}
int d,n;
int a[133][133];
int dp[133][133];
int pr1,pr2;
int main (){
d=wread();n=wread();
for ( int i(1);i<=n;++i ){
int x(wread()),y(wread()),k(wread());
a[x+1][y+1]=k;
}
for (int i(1);i<=129;++i){
for (int j(1);j<=129;++j){
dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1] + a[i][j];
}
}
for (int i(1);i<=129;++i){
for (int j(1);j<=129;++j){
int u(max(1,i-d)),d2(min(129,i+d)),l(max(1,j-d)),r(min(129,j+d));
int nx=dp[d2][r] - dp[u-1][r] - dp[d2][l-1] + dp[u-1][l-1] ;
if ( nx > pr2) pr2 = nx , pr1=1;
else if ( nx == pr2) pr1++;
}
}
printf("%d %d\n",pr1,pr2);
return 0;
}