ZOJ3578(Matrix)

Matrix

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Time Limit: 10 Seconds      Memory Limit: 131072 KB

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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！）

 #include<cstdio>
 #include<cstring>
 #include<queue>
 #include<algorithm>
 using namespace std;
 struct Matrix
 {
   int x1,x2,y1,y2,maxn;
 };
 Matrix matrix [+];
 bool cover(Matrix a,Matrix b)
 {
     if(a.x1>=b.x2||b.x1>=a.x2) return false;
     if(a.y1>=b.y2||b.y1>=a.y2) return false;
     return true;
 }
 int n,m,c;
 int x,y,h,a,b;
 int main()
 {
     while(~scanf("%d%d%d",&n,&m,&c))
     {
         int ans=;

         memset(matrix,,sizeof(matrix));
         for(int i=;i<c;i++)
         {
             int temp=;
             scanf("%d%d%d%d%d",&a,&b,&h,&x,&y);
             matrix[i].x1=x;
             matrix[i].x2=x+a;
             matrix[i].y1=y;
             matrix[i].y2=y+b;
             //matrix[i].h=h;
             for(int j=;j<i;j++)
             {
                 if(cover(matrix[i],matrix[j]))
                     temp=max(temp,matrix[j].maxn);
             }
             matrix[i].maxn=temp+h;
             ans=max(ans,matrix[i].maxn);
         }
         printf("%d\n",ans);
     }

     return ;
 }

 //1 2
 //1
 //1 5