poj1163The Triangle(动态规划)

The Triangle

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 41970		Accepted: 25383

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 5

Sample Output

30

题意：一列数，第一层一个，第二层两个，依次递推，第N层有N个。问从第一层到第N层的一条路径上使得这条路径上的数值和最大。

解题思路：显然是动态规划，在每层的一个状态d(i,j),都可以有a[i][j]+max[d(i+1,j),d(i+1,j+1)]递推得到。

自底向上，依次向上。最后d(1,1)即为最大和。

下面是自底向上的代码；同理自上向底也是可以的。其中最大的路径和在最后一行的其中一个（自上向下结果都分散了而自下向上最后汇聚第一行，第一行只有一点d（1,1）所以自下向上结果就在d(1,1)），所以用自上向下的最后要遍历一次求出最大和。

代码：

#include<stdio.h>
#include<string.h>
int max(int x,int y)
{
return x>y?x:y;
}
int main()
{
int a[120][120],dp[120][120];
int n,i,j;
while(scanf("%d",&n)!=EOF)
{
memset(a,0,sizeof(a));
memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++)
{
for(j=1;j<=i;j++)
scanf("%d",&a[i][j]);
}
for(i=n;i>=1;i--)
dp
[i]=a
[i];
for(i=n-1;i>=1;i--)
{
for(j=1;j<=i;j++)
dp[i][j]=a[i][j]+max(dp[i+1][j],dp[i+1][j+1]);
}
printf("%d\n",dp[1][1]);
}
return 0;
}

自上向下的代码：

#include<stdio.h>
#include<string.h>
int max(int x,int y)
{
return x>y?x:y;
}
int main()
{
int a[120][120],dp[120][120];
int n,i,j;
while(scanf("%d",&n)!=EOF)
{
memset(a,0,sizeof(a));
memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++)
{
for(j=1;j<=i;j++)
scanf("%d",&a[i][j]);
}
dp[1][1]=a[1][1];
for(i=2;i<=n;i++)
{
for(j=1;j<=i;j++)
{
dp[i][j]=a[i][j]+max(dp[i-1][j-1],dp[i-1][j]);
}
}
int s=-1;
for(i=1;i<=n;i++)     //遍历最后一行求出最大和；
if(dp
[i]>s)
s=dp
[i];
printf("%d\n",s);
}
return 0;
}