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Cow Bowling

2024-07-17 08:45:06   225人阅读

Cow Bowling
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 13481 Accepted: 8909

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

573 88 1 02 7 4 44 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.
数塔问题！！！又名数字三角形
 
递推方法
#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>#define N 351using namespace std;int main(){    int n,i,j,dp[N][N],map[N][N];    scanf("%d",&n);    for(i=1;i<=n;i++)       for(j=1;j<=i;j++)       {            scanf("%d",&map[i][j]);       }    for(i=1;i<=n;i++)        dp[n][i]=map[n][i];//最后一行，用新数组存好。运算时新数组其他为原始的0，计算更加好     for(i=n-1;i>=1;i--)        for(j=1;j<=i;j++)            dp[i][j]=map[i][j]+max(dp[i+1][j],dp[i+1][j+1]);//用到max()的时候，要么自己写一个max比较函数，要么用函数名 algorithm和using namespace std一起     printf("%d",dp[1][1]);    return 0;}

 


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