POJ 3176-Cow Bowling(DP||记忆化搜索) - c++编程基础

Cow Bowling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14210		Accepted: 9432

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

Sample Output

数字三角形问题。。可以自底向上坐dp dp[i][j]=ma[i][j]+max(dp[i+1][j],dp[i+1][j+1])

巨水 。。想当初半年前自己懵懵懂懂的刷dp啥都不懂。。哎  真是个悲伤的故事。。

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                 #define ll long long #define maxn 360 #define pp pair
                
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                  =0;i--) for(int j=0;j<=i;j++) dp[i][j]=max(ma[i][j]+dp[i+1][j],ma[i][j]+dp[i+1][j+1]); printf("%d\n",dp[0][0]); } int main() { while(~scanf("%d",&n)) { for(int i=0;i
                  
                 
                
               
             
            
           
          
         
        
       
      
     
    
   
也可以自顶向下记忆化搜索。。然后状态数组含义都差不多 。。个人觉着搜索比较好写。。。
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                 #define ll long long #define maxn 360 #define pp pair
                
                  #define INF 0x3f3f3f3f #define max(x,y) ( ((x) > (y)) ? (x) : (y) ) #define min(x,y) ( ((x) > (y)) ? (y) : (x) ) using namespace std; int n,dp[maxn][maxn],ma[maxn][maxn]; int dfs(int x,int y) { if(x==n-1)return ma[x][y]; if(dp[x][y]>=0)return dp[x][y]; dp[x][y]=0; dp[x][y]+=(ma[x][y]+max(dfs(x+1,y),dfs(x+1,y+1))); return dp[x][y]; } int main() { while(~scanf("%d",&n)) { for(int i=0;i