POJ 3176 Cow Bowling
Cow Bowling
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 12121 | Accepted: 7985 |
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:
Given a triangle with N (1 <= N <= 350) rows, determine the highest possible sum achievable.
7 3 8 8 1 0 2 7 4 4 4 5 2 6 5Then 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.
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 5The highest score is achievable by traversing the cows as shown above.
动态规划最基础最经典的问题
设走到第i行左数第j个点时能达到的最大值为f(i,j),该点值为num[i][j]
状态转移方程为f(i,j)=max(f(i-1,j-1),f(i-1,j))+num[i][j](注意边界处要特殊判断)
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 5 using namespace std; 6 7 int n; 8 int num[500][500],dp[500][500]; 9 10 int f(int x,int y) 11 { 12 if(dp[x][y]!=-1) 13 return dp[x][y]; 14 if(x==1) 15 return dp[x][y]=num[x][y]; 16 if(y==1) 17 return dp[x][y]=f(x-1,y)+num[x][y]; 18 if(y==x) 19 return dp[x][y]=f(x-1,y-1)+num[x][y]; 20 return dp[x][y]=max(f(x-1,y-1),f(x-1,y))+num[x][y]; 21 } 22 23 int main() 24 { 25 while(scanf("%d",&n)==1) 26 { 27 for(int i=0;i<500;i++) 28 fill(dp[i],dp[i]+500,-1); 29 for(int i=1;i<=n;i++) 30 for(int j=1;j<=i;j++) 31 scanf("%d",&num[i][j]); 32 int ans=0; 33 for(int j=1;j<=n;j++) 34 if(f(n,j)>ans) 35 ans=dp[n][j]; 36 printf("%d\n",ans); 37 } 38 39 return 0; 40 }