| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 13464 | Accepted: 8897 |
Description
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. Input
Output
Sample Input
5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5
Sample Output
30
Hint
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.简单的数塔DP问题。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int M = 350 +5;
int Cow[M][M];
int main()
{
int n;
while(scanf("%d", &n)!=EOF)
{
memset(Cow, 0, sizeof(Cow));
for(int i=1; i<=n; i++)
for(int j=1; j<=i; j++)
{
scanf("%d", &Cow[i][j]);
}
for(int i=n-1; i>=1; i--)
for(int j=1; j<=i; j++)
Cow[i][j]+=max(Cow[i+1][j], Cow[i+1][j+1]);
cout<<Cow[1][1]<<endl;
}
return 0;
}
POJ 3176：Cow Bowling,布布扣,bubuko.com
原文：http://blog.csdn.net/u013487051/article/details/37344035