poj 3176（简单动态规划）

最新推荐文章于 2016-04-21 18:33:40 发布

原创最新推荐文章于 2016-04-21 18:33:40 发布 · 653 阅读

0 ·

CC 4.0 BY-SA版权

动态规划专栏收录该内容

15 篇文章

订阅专栏

本文探讨了在CowBowling游戏中如何通过优化路径选择实现最大得分的策略，并提供了一个高效的算法来解决该问题。

Cow Bowling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13262		Accepted: 8759

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

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.

f[i][j]表示到第i行j列时的最大值

直接看AC代码吧

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int f[360][360];
int a[360][360];
int main(){
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    while(scanf("%d",&n)!=EOF){
        for(int i=1;i<=n;i++)
            for(int j=1;j<=i;j++)
                scanf("%d",&a[i][j]);
        memset(f,0,sizeof(f));
        f[1][1]=a[1][1];
        for(int i=2;i<=n;i++)
            for(int j=1;j<=i;j++){
                if(j==1)
                    f[i][j]=f[i-1][j]+a[i][j];
                else
                    f[i][j]=max(f[i-1][j],f[i-1][j-1])+a[i][j];
            }
        int ans=0;
        for(int i=1;i<=n;i++)
            ans=max(ans,f[n][i]);
        printf("%d\n",ans);
    }
    return 0;
}