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
1 /* 2 ID:fffgrdcc1 3 PROB:numtri 4 LANG:C++ 5 */ 6 #include<cstdio> 7 #include<iostream> 8 #include<cstring> 9 using namespace std; 10 int a[1002][1002],f[1002][1002]; 11 int main() 12 { 13 freopen("numtri.in","r",stdin); 14 freopen("numtri.out","w",stdout); 15 memset(f,0,sizeof(f)); 16 memset(a,0,sizeof(a)); 17 int n; 18 scanf("%d",&n); 19 for(int i=1;i<=n;i++) 20 { 21 for(int j=1;j<=i;j++) 22 { 23 scanf("%d",&a[i][j]); 24 } 25 } 26 for(int i=n+1;i>1;i--) 27 { 28 for(int j=1;j<=i;j++) 29 { 30 f[i-1][j]=max(f[i-1][j],f[i][j]+a[i-1][j]); 31 f[i-1][j-1]=max(f[i-1][j-1],f[i][j]+a[i-1][j-1]); 32 } 33 } 34 printf("%d ",f[1][1]); 35 return 0; 36 }
（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）