POJ 3176 Cow Bowling

Cow Bowling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12121		Accepted: 7985

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.

 

动态规划最基础最经典的问题

设走到第i行左数第j个点时能达到的最大值为f(i,j)，该点值为num[i][j]

状态转移方程为f(i,j)=max(f(i-1,j-1),f(i-1,j))+num[i][j]（注意边界处要特殊判断）

 

 

#include<iostream>
#include<cstdio>
#include<algorithm>

using namespace std;

int n;
int num[500][500],dp[500][500];

int f(int x,int y)
{
    if(dp[x][y]!=-1)
        return dp[x][y];
    if(x==1)
        return dp[x][y]=num[x][y];
    if(y==1)
        return dp[x][y]=f(x-1,y)+num[x][y];
    if(y==x)
        return dp[x][y]=f(x-1,y-1)+num[x][y];
    return dp[x][y]=max(f(x-1,y-1),f(x-1,y))+num[x][y];
}

int main()
{
    while(scanf("%d",&n)==1)
    {
        for(int i=0;i<500;i++)
            fill(dp[i],dp[i]+500,-1);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=i;j++)
                scanf("%d",&num[i][j]);
        int ans=0;
        for(int j=1;j<=n;j++)
            if(f(n,j)>ans)
                ans=dp[n][j];
        printf("%d
",ans);
    }

    return 0;
}