P1004 and p1004 squares

P1004 and p1004 squares
Description

The square plot with N * N (N <= 9) is filled with positive integers in some squares, while in other squares

The number of people is 0. As shown in (see example ):

A 0  0  0  0  0  0  0  0 0  0 13  0  0  6  0  0 0  0  0  0  7  0  0  0 0  0  0 14  0  0  0  0 0 21  0  0  0  4  0  0 0  0 15  0  0  0  0  0 0 14  0  0  0  0  0  0 0  0  0  0  0  0  0  0.                       B

Someone starts from the point in the upper left corner of the graph and can walk down or right until B in the lower right corner.

Point. On the way he walked, he could take away the number from the square (the square after the square is removed will become a number 0 ).

This person takes two steps from A.M. To a.m. and tries to find two such paths to maximize the sum of the obtained numbers.

Input/Output Format Input Format:

The first line of the input is an integer N (representing the square map of N * N). The next line has three integers, the first two

Represents the position. The third number is the number placed on the position. A single row of 0 indicates that the input is complete.

Output Format:

Only one integer is output, indicating the maximum sum obtained from the two paths.

Input and Output sample Input example #1:

82 3 132 6  63 5  74 4 145 2 215 6  46 3 157 2 140 0  0

Output sample #1:

67

Description

NOIP 2000 raise group fourth question

 

There are four scenarios:
① Both roads reach this point from the top

② Both roads arrive at this point from the left

③ The first route arrives at this point from the left and the second route arrives at this point.

④ The first route arrives at this point from the top, and the second route arrives at this point on the left.

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 using namespace std; 6 void read(int & n) 7 { 8     char c='+';int x=0;     9     while(c<'0'||c>'9')c=getchar();10     while(c>='0'&&c<='9')11     {12         x=x*10+c-48;13         c=getchar();14     }15     n=x;16 }17 const int MAXN=10;18 int a[MAXN][MAXN];19 int dp[MAXN][MAXN][MAXN][MAXN];20 int main()21 {22     //ios::sync_with_stdio(false);23     int n;24     read(n);25     while(1)26     {27         int x,y,z;28         read(x);read(y);read(z);29         if(x==0&&y==0&&z==0)30         break;31         a[x][y]=z;32     }33         34 35     for(int i=1;i<=n;i++)36         for(int j=1;j<=n;j++)37             for(int k=1;k<=n;k++)38                 for(int l=1;l<=n;l++)39                 {40                     if(i+j!=k+l)continue;41                     dp[i][j][k][l]=max(max(dp[i-1][j][k-1][l],42                                        dp[i-1][j][k][l-1]),43                                        max(dp[i][j-1][k-1][l],44                                        dp[i][j-1][k][l-1]))+a[i][j]+a[k][l];45                     if(i==k&&j==l)46                     dp[i][j][k][l]-=a[i][j];47                 }48     printf("%d",dp[n][n][n][n]);49     return 0;50 }