USACO1.5.1--Number Triangles

Number Triangles

Consider the number triangle shown below. Write a program that calculates the highest sum of numbers that can be 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.

          7



        3   8



      8   1   0



    2   7   4   4



  4   5   2   6   5

In the sample above, the route from 7 to 3 to 8 to 7 to 5 produces the highest sum: 30.

PROGRAM NAME: numtri

INPUT FORMAT

The first line contains R (1 <= R <= 1000), the number of rows. Each subsequent line contains the integers for that particular row of the triangle. All the supplied integers are non-negative and no larger than 100.

SAMPLE INPUT (file numtri.in)

5

7

3 8

8 1 0

2 7 4 4

4 5 2 6 5

OUTPUT FORMAT

A single line containing the largest sum using the traversal specified.

SAMPLE OUTPUT (file numtri.out)

30
解题思路：经典的DP问题。状态转移方程：f[i][j]=a[i][j]+max(f[i-1,j-1],f[i-1,j])。把数组开成局部变量了，然后IDE一运行就中断结束了。题解到USACO上居然AC了。然后纠结了好久才找到原因。以后记得大数组开成全局的。

View Code

 1 /*

 2 ID:spcjv51

 3 PROG:numtri

 4 LANG:C

 5 */

 6 #include<stdio.h>

 7 int a[1005][1005],f[1005][1005];

 8 int max(int a,int b)

 9 {

10     return(a>b?a:b);

11 }

12 int main(void)

13 {

14     freopen("numtri.in","r",stdin);

15     freopen("numtri.out","w",stdout);

16     int i,j,n,ans;

17     scanf("%d",&n);

18     memset(f,0,sizeof(f));

19     for(i=1; i<=n; i++)

20         for(j=1; j<=i; j++)

21             scanf("%d",&a[i][j]);

22     f[1][1]=a[1][1];

23     for(i=2; i<=n; i++)

24         for(j=1; j<=i; j++)

25             f[i][j]=a[i][j]+max(f[i-1][j],f[i-1][j-1]);

26     ans=-1;

27     for(j=1; j<=n; j++)

28         if(f[n][j]>ans) ans=f[n][j];

29     printf("%d\n",ans);

30     return 0;

31 }