poj 3176 动态规划

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Cow Bowling

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 19704 Accepted: 13056

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.

我们可以从下往上推，每一步都取下面相邻的两个值中的较大的值，推到最后就是答案

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

#include<iostream>#include<string.h>using namespace std;int map[351][351];int n;int main(){    while(cin>>n)    {        memset(map,0,sizeof(map));        for(int i=1;i<=n;i++)        {            for(int j=1;j<=i;j++)                cin>>map[i][j];        }        for(int i=n-1;i>=1;i--)        {            for(int j=1;j<=i;j++)            {            map[i][j]=max(map[i+1][j],map[i+1][j+1])+map[i][j];            }        }        cout<<map[1][1]<<endl;    }    return 0;}