[HNOI2003]激光炸弹

嘟嘟嘟

因为数据只有5000，所以可以O(n²)暴力。

首先预处理二维前缀和，然后枚举正方形的左上角，将每一次的得到的总价值去最大，作为答案。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<cstdlib>
 7 #include<vector>
 8 #include<queue>
 9 #include<stack>
10 #include<cctype>
11 using namespace std;
12 #define enter puts("")
13 #define space putchar(' ')
14 #define Mem(a) memset(a, 0, sizeof(a))
15 typedef long long ll;
16 typedef double db;
17 const int INF = 0x3f3f3f3f;
18 const db eps = 1e-8;
19 const int maxn = 5e3 + 1;
20 inline ll read()
21 {
22     ll ans = 0;
23     char ch = getchar(), last = ' ';
24     while(!isdigit(ch)) {last = ch; ch = getchar();}
25     while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
26     if(last == '-') ans = -ans;
27     return ans;
28 }
29 inline void write(ll x)
30 {
31     if(x < 0) putchar('-'), x = -x;
32     if(x >= 10) write(x / 10);
33     putchar(x % 10 + '0');
34 }
35 
36 int n, r;
37 int sum[maxn + 5][maxn + 5], ans = 0;
38 
39 int main()
40 {
41     n = read(); r = read();
42     for(int i = 1; i <= n; ++i) sum[read() + 1][read() + 1] = read();    //这个操作有点秀 
43     for(int i = 1; i <= maxn; ++i)
44         for(int j = 1; j <= maxn; ++j) sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + sum[i][j];
45     for(int i = 1; i <= maxn - r + 1; ++i)
46         for(int j = 1; j <= maxn - r + 1; ++j) 
47             ans = max(ans, sum[i + r - 1][j + r - 1] - sum[i + r - 1][j - 1] - sum[i - 1][j + r - 1] + sum[i - 1][j - 1]);
48     write(ans); enter;
49     return 0;
50 }

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