E - Evaluate Matrix Sum

最新推荐文章于 2022-11-25 18:45:13 发布

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最新推荐文章于 2022-11-25 18:45:13 发布

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原文链接：http://www.cnblogs.com/zhangying/p/3924337.html

本文介绍了一种高效的矩阵子区域求和算法，通过预处理矩阵元素并利用累积和技巧来快速计算任意指定子区域内的元素平方和。适用于处理大量查询的情况。

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Description

Given a matrix, the elements of which are all integer number from 0 to 50, you are required to evaluate the square sum of its specified sub-matrix.

Input

The first line of the input contains a single integer T (1 <= T <= 5), the number of test cases.

For each test case, the first line contains two integers m and n (1 <= m, n <= 500), which are the row and column sizes of the matrix, respectively. The next m lines with n numbers each gives the elements of the matrix.

The next line contains a single integer N (1 <= N <= 100,000), the number of queries. The next N lines give one query on each line, with four integers r1, c1, r2, c2 (1 <= r1 <= r2 <= m, 1 <= c1 <= c2 <= n), which are the indices of the upper-left corner and lower-right corner of the sub-matrix in question.

Output

For each test case, first print the number of the test case, then N lines with one number on each line, the required square sum. Refer to the sample output for details.

Sample Input

Sample Output

Case 1:
46
45
Case 2:
193

题意：
给一个n*m矩阵  下表（1,1)~~~(n,m)
给两个点做对角线   求中间矩阵的值
朴素算法也可以过   但是这个在不断输入的同时也将值存在数组中   查询更快

  #include <iostream>
 #include <string.h>
 #include <stdio.h>

 using namespace std;

 int main()
 {
     int t;
     int n,m;
     int k;
     int tt,x1,x2,y1,y2;
     int num;
     int map[505][505];
     scanf("%d",&t);
     for(num=1;num<=t;num++)
     {
         scanf("%d%d",&n,&m);
         memset(map,0,sizeof(map));
         for(int i=1;i<=n;i++)
         {
             for(int j=1;j<=m;j++)
             {
                 scanf("%d",&k);
                 map[i][j]=k*k+map[i-1][j]+map[i][j-1]-map[i-1][j-1];
             }
         }
         printf("Case %d:\n",num);
         scanf("%d",&tt);
         while(tt--)
         {
             scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
             int ans=map[x1-1][y1-1]+map[x2][y2]-map[x1-1][y2]-map[x2][y1-1];
             printf("%d\n",ans);

         }
     }
     return 0;
 }
 
 
 
 
 
 
 
 #include<iostream>
#include <string.h>
#include <stdio.h>

using namespace std;

int a[ 510 ][ 510 ];
int sum[ 510 ][ 510 ];

int main()
{
    int t,n,m,tt;
    int ss;
    int k;
    int r1,c1,r2,c2;
    scanf("%d",&t);
    for(k=1; k<=t; k++)
    {
        scanf("%d%d",&n,&m);
        memset(sum,0,sizeof(sum));
        for(int i=1; i<=n; i++)
        {
            ss = 0;
            for(int j=1; j<=m; j++)
            {
                scanf("%d",&a[i][j]);
                ss += a[i][j] * a[i][j];
                sum[i][j] = sum[ i - 1 ][ j ] + ss;
            }
        }
        printf("Case %d:\n",k);
        scanf("%d",&tt);
        while(tt--)
        {
            scanf("%d%d%d%d",&r1,&c1,&r2,&c2);
            printf("%d\n",sum[ r2 ][ c2 ] - sum[ r1-1 ][ c2 ] -  sum[ r2 ][ c1 - 1 ] + sum[ r1 - 1 ][ c1 - 1 ]  );
        }
    }
    return 0;

}