Cow Bowling
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 16448 Accepted: 10957
Description
The cows don’t use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard bowling-pin-like triangle like this:
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
Then the other cows traverse the triangle starting from its tip and moving “down” to one of the two diagonally adjacent cows until the “bottom” row is reached. The cow’s score is the sum of the numbers of the cows visited along the way. The cow with the highest score wins that frame.
Given a triangle with N (1 <= N <= 350) rows, determine the highest possible sum achievable.
Input
Line 1: A single integer, N
Lines 2..N+1: Line i+1 contains i space-separated integers that represent row i of the triangle.
Output
Line 1: The largest sum achievable using the traversal rules
Sample Input
5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
Sample Output
30
Hint
Explanation of the sample:
7
*
3 8
*
8 1 0
*
2 7 4 4
*
4 5 2 6 5
The highest score is achievable by traversing the cows as shown above.
题意：一个n层的三角形，从第一行第一列开始向下走，每次只能向下一行且只能走到与之到相邻的两块，求到最后一行路程的最大值。
解题思路：动规
f[i][j]表示到第i行第j列路程的最大值
走到第i行第j列的路程只能与走到第i-1行第j列和第i-1行第i-1列有关
不难写出状态转移方程为：f[i][j]=max(f[i][j]+f[i-1][j-1],f[i][j]+f[i-1][j]);
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int n,f[666][666];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=i;j++)
{
scanf("%d",&f[i][j]);
}
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=i;j++)
{
f[i][j]=max(f[i][j]+f[i-1][j-1],f[i][j]+f[i-1][j]);
}
}
for(int i=1;i<=n;i++)
{
if(f[n][i]>f[0][0]) f[0][0]=f[n][i];
}
printf("%d",f[0][0]);
}