BZOJ 1218: [HNOI2003] laser bombs (prefix and two-dimensional)

1218: [HNOI2003] laser bombs

Description

A novel laser bombs that can destroy all of the length of one side of the square R of the target. There are map n (N <= 10000) target, an integer Xi, Yi (its value in [0,5000]) indicates the position of the target on a map, each has a certain value. Laser bombs served by satellite positioning, but it has a disadvantage that its explosion range, i.e., the side length of the sides of the square must R and x, y-axis. If the target is located on the edge of the square blasting, the target will not be destroyed. 

Input

The first line of the input file and the integer n a positive integer R, the next n lines each have three positive integers, respectively, xi, yi, vi

Output

The output file is only a positive integer representing up to blow up a bomb on a map total value of the number of goals (results not more than 32,767).

Sample Input

2 1

0 0 1

1 1 1

Sample Output

1

 

Ideas: a matrix product refresh most value like two-dimensional prefix and enumeration ......

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #include<map>
 7 #include<set>
 8 #include<vector>
 9 #include<queue>
10 #include<stack>
11 #include<list>
12 //#include<unordered_map>
13 using namespace std;
14 #define ll long long 
15 const int mod=1e9+7;
16 const int inf=1e9+7;
17  
18 const int maxn=5e3;
19  
20 int sum[maxn+10][maxn+10]; 
21  
22 int main()
23 {
24     ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
25      
26     int n,r;
27      
28     cin>>n>>r;
29     int x,y,w;
30      
31     int mn=-1;
32     for(int i=0;i<n;i++)
33     {
34         cin>>x>>y>>w;
35         x++;
36         y++;
37         mn=max(mn,x);
38         mn=max(mn,y);
39         sum[x][y]=w;
40     }
41      
42     for(int i=1;i<=mn;i++)
43     {
44         for(int j=1;j<=mn;j++)
45         {
46             sum[i][j]+=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1];
47         }
48     }
49      
50     int maxx=-1;
51      
52     for(int i=r;i<=mn;i++)
53     {
54         for(int j=r;j<=mn;j++)
55         {
56             maxx=max(maxx,sum[i][j]-sum[i-r][j]-sum[i][j-r]+sum[i-r][j-r]);
57         }
58     }
59      
60     cout<<maxx<<endl;
61      
62     return 0;
63 }

 

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Origin www.cnblogs.com/xwl3109377858/p/11286039.html