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。

/*
ID:spcjv51
PROG:numtri
LANG:C
*/
#include<stdio.h>
int a[1005][1005],f[1005][1005];
int max(int a,int b)
{
    return(a>b?a:b);
}
int main(void)
{
    freopen("numtri.in","r",stdin);
    freopen("numtri.out","w",stdout);
    int i,j,n,ans;
    scanf("%d",&n);
    memset(f,0,sizeof(f));
    for(i=1; i<=n; i++)
        for(j=1; j<=i; j++)
            scanf("%d",&a[i][j]);
    f[1][1]=a[1][1];
    for(i=2; i<=n; i++)
        for(j=1; j<=i; j++)
            f[i][j]=a[i][j]+max(f[i-1][j],f[i-1][j-1]);
    ans=-1;
    for(j=1; j<=n; j++)
        if(f[n][j]>ans) ans=f[n][j];
    printf("%d\n",ans);
    return 0;
}