ZOJ3578(Matrix)
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
