【坐标型动态规划】Number Triangles数字金字塔(Usaco_Training 1.5)

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

 

很经典的坐标型动规的题目了，f[i][j]表示当前坐标所能获得的最大值

当前坐标只能由正上面和作上面的坐标得到，所以

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

并且不用判断坐标是否有效（无效的坐标为0，对结果没有任何影响）

 

还有这里抱怨一下USACO，同一个程序连交了3次！！！第一次居然第一组数据就超时，第二次第六组数据超时，第三次就过了，什么东西啊，一个程序的误差也不能这么大吧！！！

C++ Code

 

/*
ID: jiangzh15
TASK: numtri
LANG: C++ http://blog.csdn.net/jiangzh7 */
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAXN 1010

int n,f[MAXN][MAXN];
int map[MAXN][MAXN];

int main()
{
freopen("numtri.in","r",stdin);
freopen("numtri.out","w",stdout);
scanf("%d",&n);
for(int i=1;i<=n;i++)
for(int j=1;j<=i;j++)
scanf("%d",&map[i][j]);
for(int i=1;i<=n;i++)
for(int j=1;j<=i;j++)
f[i][j]=max(f[i-1][j],f[i-1][j-1])+map[i][j];
int maxx=f
[1];
for(int i=1;i<=n;i++) maxx=max(maxx,f
[i]);
printf("%d\n",maxx);
return 0;
}