[dp]poj3176 Cow Bowling

Cow Bowling

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 15103 Accepted: 10047

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.

题目意思很简单，就是给一个三角形数列，从第一步开始，可以往左下或者右下走，求出从上到下取出的一条最大权值。

定义 d[i][j]  为从节点(i,j)开始到最底层的最大权值

状态转移方程： dp[i][j] += max(dp[i+1][j],dp[i+1][j+1]);

有了这个方程，题目就很好写了。

#include <cstdio>#include <iostream>#include <cstring>using namespace std;const int MAXN = 400;int dp[MAXN][MAXN];int main(){    int N;    while(cin>>N)    {        memset(dp,0,sizeof(dp));        for(int i = 1;i<=N;i++)            for(int j = 1; j<=i;j++)                scanf("%d",&dp[i][j]);        for(int i = N-1;i>=1;i--)            for(int j = 1; j <= i;j++)                dp[i][j] += max(dp[i+1][j],dp[i+1][j+1]);        cout<<dp[1][1]<<endl;    }}