poj-3176-Cow Bowling

Cow Bowling

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 18724 Accepted: 12468

---

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.

题意：共有n行，第i行有i个数字，求第n层时找到权值和最大的路线，其中每个数字只能与它上面或左上方的数字相连。

具体看代码：

#include <stdio.h>#include <string.h>#define max(a,b) a>b?a:bint main(){    int n,i,j;    int a[355][355],dp[355][355];//表示以第i行j列的位置作为终点的路线的最大权值    memset(a,0,sizeof(a));    memset(dp,0,sizeof(dp));    scanf("%d",&n);    for(i=1;i<=n;i++){        for(j=1;j<=i;j++){            scanf("%d",&a[i][j]);        }    }    for(i=1;i<=n;i++){        for(j=1;j<=i;j++)        {            dp[i][j]=max(dp[i-1][j]+a[i][j],dp[i-1][j-1]+a[i][j]);        } //找到每个位置权值的最大值    }    int max=0;    for(i=1;i<=n;i++){        max=max(max,dp[n][i]);    }//找到最后一行的最大值    printf("%d\n",max);    return 0;}