USACO 1.5 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

题目大意：数字三角形，动态规划（DP）入门题目，说的是有一个一群数字排列成三角形，你现在站在上顶点，现在想要走到底边，每次你可以选择走左下或是右下，每次踩到的数字都累加起来，求最大累加和。
思路：这题搜索不好搞，复杂度2^1000简直开玩笑，仔细思索，发现如果我知道脚下的两个点哪个更优，我就有唯一的行走策略了，那就从下往上推，f[i][j]表示从这点出发到达底边的最大累加和，他可以更新它左上的和右上的，从下往上推一遍答案就出来了，就是f[1][1];代码见下面

/*
ID:fffgrdcc1
PROB:numtri
LANG:C++
*/
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int a[1002][1002],f[1002][1002];
int main()
{
    freopen("numtri.in","r",stdin);
    freopen("numtri.out","w",stdout);
    memset(f,0,sizeof(f));
    memset(a,0,sizeof(a));
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=i;j++)
        {
            scanf("%d",&a[i][j]);
        }
    }
    for(int i=n+1;i>1;i--)
    {
        for(int j=1;j<=i;j++)
        {
            f[i-1][j]=max(f[i-1][j],f[i][j]+a[i-1][j]);
            f[i-1][j-1]=max(f[i-1][j-1],f[i][j]+a[i-1][j-1]);
        }
    }
    printf("%d
",f[1][1]);
    return 0;
}

（这题原来写过，所以写的略快，边界的处理比较粗糙，不过因为我提前把边界外一层留了出来且都设为0，所以不会影响答案）