poj1163 - DP递推递归写法

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The Triangle
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 50977   Accepted: 30888

Description

7 
3 8
8 1 0
2 7 4 4
4 5 2 6 5
(Figure 1)
Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right. 

Input

Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.

Output

Your program is to write to standard output. The highest sum is written as an integer.

Sample Input

5
7
3 8
8 1 0 
2 7 4 4
4 5 2 6 5

Sample Output

30

Source

 
DP的递推写法
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 #include <iostream>
 5 using namespace std;
 6 
 7 int dp[105][105];
 8 int f[105][105];
 9 
10 int func(int n)
11 {
12     for(int i=n-1;i>=1;i--)
13     {
14         for(int j=1;j<=i;j++)
15         {
16             dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])+f[i][j];
17         }
18     }
19     return dp[1][1];
20 }
21 int main()
22 {
23     int n;
24     while(~scanf("%d",&n))
25     {
26         for(int i=1;i<=n;i++)
27         {
28             for(int j=1;j<=i;j++)
29             {
30                 scanf("%d",&f[i][j]);
31             }
32         }
33         for(int i=1;i<=n;i++) dp[n][i]=f[n][i];
34         printf("%d\n",func(n));
35     }
36     
37     return 0;
38 }


DP的递归写法

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 #include <iostream>
 5 using namespace std;
 6 
 7 int dp[105][105];
 8 int f[105][105];
 9 int n;
10 
11 int func(int i,int j)
12 {
13     if(i==n) return dp[i][j]=f[i][j];
14     if(dp[i][j]!=-1) return dp[i][j];
15     return dp[i][j]=max(func(i+1,j),func(i+1,j+1))+f[i][j];
16         
17 }
18 int main()
19 {
20     
21     while(~scanf("%d",&n))
22     {
23         memset(dp,-1,sizeof(dp));
24         for(int i=1;i<=n;i++)
25         {
26             for(int j=1;j<=i;j++)
27             {
28                 scanf("%d",&f[i][j]);
29             }
30         }
31         
32         printf("%d\n",func(1,1));
33     }
34     
35     return 0;
36 }
 

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