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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 14748 | Accepted: 9804 |
Description
7Then 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.
3 8
8 1 0
2 7 4 4
4 5 2 6 5
Input
Output
Sample Input
5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5
Sample Output
30
Hint
7The highest score is achievable by traversing the cows as shown above.
*
3 8
*
8 1 0
*
2 7 4 4
*
4 5 2 6 5
Source
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define maxn 351 int dp[maxn][maxn]; int a[maxn][maxn]; int n; void solve() { for(int i=n-1;i>=0;i--) { for(int j=0;j<=i;j++) { dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])+a[i][j]; } } printf("%d\n",dp[0][0]); } int main() { while(scanf("%d",&n)!=EOF) { for(int i=0;i<n;i++) { for(int j=0;j<=i;j++) { scanf("%d",&a[i][j]); } } memset(dp,0,sizeof(dp)); solve(); } return 0; }
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原文地址：http://www.cnblogs.com/qianyanwanyu--/p/4375132.html
