[题目链接]
[算法]
二维前缀和
[代码]
#include<bits/stdc++.h> using namespace std; #define MAXN 300 #define calc(x1,y1,x2,y2) s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1] int i,j,d,n,ans,cnt,x,y,k,ta,tb,tx,ty; int s[MAXN][MAXN]; template <typename T> inline void read(T &x) { int f = 1; x = 0; char c = getchar(); for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; } for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0'; x *= f; } int main() { read(d); read(n); for (i = 1; i <= n; i++) { read(x); read(y); read(k); x++; y++; s[x][y] = k; } for (i = 1; i <= 129; i++) { for (j = 1; j <= 129; j++) { s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + s[i][j]; } } for (i = 1; i <= 129; i++) { for (j = 1; j <= 129; j++) { ta = max(i - d,1); tb = max(j - d,1); tx = min(i + d,129); ty = min(j + d,129); if (calc(ta,tb,tx,ty) > ans) { ans = calc(ta,tb,tx,ty); cnt = 1; } else if (calc(ta,tb,tx,ty) == ans) cnt++; } } printf("%d %d ",cnt,ans); return 0; }