poj 3176 Cow Bowling动态规划

Cow Bowling

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12462 Accepted: 8219

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.
这道题意思很简单就是给你一个三角形的矩阵[i][j]只能走向自己的下面[i+1][j]和右下面[i+1][j+1]并且权值相加
求走完所有的行最大的权值，这道题同样可以通过我的吃豆机器人解释吃豆机器人链接
很好的解释了状态方程的由来

#include<stdio.h>#include<iostream>#include<cstring>using namespace std;int dp[355][355];int max(int a,int b){return a>b?a:b;}int main(){int i,j,n,num;while(scanf("%d",&n)!=EOF){memset(dp,0,sizeof(dp));num=0;for(i=1;i<=n;i++)for(j=1;j<=i;j++)scanf("%d",&dp[i][j]);for(i=1;i<=n;i++)for(j=1;j<=i;j++){dp[i][j]+=max(dp[i-1][j-1],dp[i-1][j]);if(dp[i][j]>num)num=dp[i][j];}printf("%d\n",num);}return 0;}

另一道题poj1163也是这个思路，代码不用改都能通过点击打开链接