Filthy Rich

They say that in Phrygia, the streets are paved with gold. You’re currently on vacation in Phrygia, and to your astonishment you discover that this is to be taken literally: small heaps of gold are distributed throughout the city. On a certain day, the Phrygians even allow all the tourists to collect as much gold as they can in a limited rectangular area. As it happens, this day is tomorrow, and you decide to become filthy rich on this day. All the other tourists decided the same however, so it’s going to get crowded. Thus, you only have one chance to cross the field. What is the best way to do so? 

Given a rectangular map and amounts of gold on every field, determine the maximum amount of gold you can collect when starting in the upper left corner of the map and moving to the adjacent field in the east, south, or south-east in each step, until you end up in the lower right corner. 

Input

The input starts with a line containing a single integer, the number of test cases. 
Each test case starts with a line, containing the two integers r and c, separated by a space (1 <= r, c <= 1000). This line is followed by r rows, each containing c many integers, separated by a space. These integers tell you how much gold is on each field. The amount of gold never negative. 
The maximum amount of gold will always fit in an int.

Output

For each test case, write a line containing “Scenario #i:”, where i is the number of the test case, followed by a line containing the maximum amount of gold you can collect in this test case. Finish each test case with an empty line. 

Sample Input

1
3 4
1 10 8 8
0 0 1 8
0 27 0 4

Sample Output

Scenario #1:
42

题意：从坐上走到右下每走一个格子就加上格子里的数，可以向右向下，和斜下走，问怎样走才使加到的数最多。

思路：本来我用的深搜，简单粗暴但是超时，后来请教了大佬，大佬给我讲了一种方法，就是每走一步都要比，先把第一行第一列加上去，第一行或第一列第二个位置对应值是它前面的值加上它本身的值，依次类推，然后开始输剩余几行的值，每输一个都要比较是从这个位置左边走过来较大还是从它上面走过来较大，确定较大的然后给这个位置赋值，那么到最后走到右下角那个位置肯定是加到了最大的。代码如下：

#include<stdio.h>
#include<string.h>
int book[1010][1010];
int main()
{
    int T,m,n,i,j,k=1;
    scanf("%d",&T);
    while(T--)
    {   memset(book,0,sizeof(book));
        int t;
        scanf("%d%d",&n,&m);
        scanf("%d",&book[0][0]);
        for(i=1;i<m;i++)
        {
            scanf("%d",&t);
            book[0][i]=book[0][i-1]+t;  //横着加第一排
        }
        for(i=1;i<n;i++)
        {
            scanf("%d",&t);
            book[i][0]=book[i-1][0]+t;//给第一列赋个值
            for(j=1;j<m;j++)
            {
                scanf("%d",&t);
                if(book[i-1][j]>book[i][j-1])
                    book[i][j]=book[i-1][j]+t;//择优走走到右下角就是最优的结果
                else
                    book[i][j]=book[i][j-1]+t;
            }
        }
        printf("Scenario #%d:\n%d\n\n",k++,book[n-1][m-1]);
    }
    return 0;
}