Matrix
Time Limit: 10 Seconds Memory Limit: 131072 KB
A N*M coordinate plane ((0, 0)~(n, m)). Initially the value of all N*M grids are 0.
An operation T(a, b, h, x, y) is defined as follow:
1. Select the maximum value in the matrix (x, y) ~ (x+a, y+b), suppose the maximum value is max
2. Change all the value in the matrix (x, y) ~ (x+a, y+b) into max+h
After C operations, please output the maximum value in the whole N*M coordinate.
Input
The input consists of several cases.
For each case, the first line consists of three positive integers N , M and C (N ≤ 1000, M ≤ 1000, C ≤ 1000). In the following C lines, each line consists of 5 non-negative number, ai, bi, hi, xi, yi (0 ≤ hi ≤ 10000, 0 ≤ xi < n, 0 ≤ yi < m).
Output
For each case, output the maximum height.
Sample Input
3 2 2 2 1 9 1 1 1 1 2 2 1
Sample Output
11
感受：判断两个矩阵是否相交，考虑一定不相交的情况。若考虑相交条件则是有可能相交。并且呀，不相交对立面不一定是相交啊（好人的对立面不一定是坏人啊，有可能是不好不坏的人啦，世界并不是由二元性组成的呀！）
1 #include<cstdio>
2 #include<cstring>
3 #include<queue>
4 #include<algorithm>
5 using namespace std;
6 struct Matrix
7 {
8 int x1,x2,y1,y2,maxn;
9 };
10 Matrix matrix [1000+5];
11 bool cover(Matrix a,Matrix b)
12 {
13 if(a.x1>=b.x2||b.x1>=a.x2) return false;
14 if(a.y1>=b.y2||b.y1>=a.y2) return false;
15 return true;
16 }
17 int n,m,c;
18 int x,y,h,a,b;
19 int main()
20 {
21 while(~scanf("%d%d%d",&n,&m,&c))
22 {
23 int ans=0;
24
25 memset(matrix,0,sizeof(matrix));
26 for(int i=0;i<c;i++)
27 {
28 int temp=0;
29 scanf("%d%d%d%d%d",&a,&b,&h,&x,&y);
30 matrix[i].x1=x;
31 matrix[i].x2=x+a;
32 matrix[i].y1=y;
33 matrix[i].y2=y+b;
34 //matrix[i].h=h;
35 for(int j=0;j<i;j++)
36 {
37 if(cover(matrix[i],matrix[j]))
38 temp=max(temp,matrix[j].maxn);
39 }
40 matrix[i].maxn=temp+h;
41 ans=max(ans,matrix[i].maxn);
42 }
43 printf("%d\n",ans);
44 }
45
46
47 return 0;
48 }
49
50 //1 2
51 //1
52 //1 5