POJ 1163 The Triangle【dp+杨辉三角加强版(递归)】
POJ 1163 The Triangle【dp+杨辉三角加强版(递归)】
The Triangle
Time Limit: 1000MS | Memory Limit: 10000K | |
|---|---|---|
Total Submissions: 49955 | Accepted: 30177 |
Description
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
(Figure 1)Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers 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.
Input
Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.
Output
Your program is to write to standard output. The highest sum is written as an integer.
Sample Input
5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5Sample Output
30Source
题目链接：http://poj.org/problem?id=1163
题目要求：
输入一个n层的三角形，第i层有i个数，求从第1层到第n层的所有路线中，权值之和最大的路线。 规定：第i层的某个数只能连线走到第i+1层中与它位置相邻的两个数中的一个。 题目分析：
典型的动态规划。
因此我们可以从下往上推，相邻的两个数中找较大的与上层相加，得出的结果相邻的两个数中再找较大的与上层相加，以此类推。用二维数组d_[][]记录从下到该点的最大值。
核心代码
d[i][j] += d[i+1][j] > d[i+1][j+1] ? d[i+1][j] : d[i+1][j+1];
最后的结果就是d[0][0]。
下面给出AC代码:
1 #include <stdio.h>
2 #include <string.h>
3 int max(int a,int b)
4 {
5 return a>b?a:b;
6 }
7 inline int read()
8 {
9 int x=0,f=1;
10 char ch=getchar();
11 while(ch<'0'||ch>'9')
12 {
13 if(ch=='-')
14 f=-1;
15 ch=getchar();
16 }
17 while(ch>='0'&&ch<='9')
18 {
19 x=x*10+ch-'0';
20 ch=getchar();
21 }
22 return x*f;
23 }
24 inline void write(int x)
25 {
26 if(x<0)
27 {
28 putchar('-');
29 x=-x;
30 }
31 if(x>9)
32 {
33 write(x/10);
34 }
35 putchar(x%10+'0');
36 }
37 int a[101][101],d[101][101];
38 int n;
39 int dp(int i,int j)
40 {
41 if(d[i][j]>=0)
42 return d[i][j];
43 return d[i][j]=a[i][j]+(i==n-1?0:max(dp(i+1,j),dp(i+1,j+1)));
44 }
45 int main()
46 {
47 int i,j;
48 while(scanf("%d",&n)!=EOF)
49 {
50 for(int i=0;i<n;i++)
51 for(j=0;j<i+1;j++)
52 scanf("%d",&a[i][j]);
53 memset(d,-1,sizeof(d));
54 printf("%d\n",dp(0,0));
55 }
56 return 0;
57 }