标签:dp
题目链接:http://poj.org/problem?id=1163
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 39022 | Accepted: 23430 |
Description
7 3 8 8 1 0 2 7 4 4 4 5 2 6 5 (Figure 1)
Input
Output
Sample Input
5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5
Sample Output
30
Source
题意;数字三角形→_→题解: DP
AC代码:
#include<iostream> #include<cstring> #define N 105 using namespace std; int dp[N][N],n; int main() { cin.sync_with_stdio(false); while(cin>>n){ memset(dp,0,sizeof(dp)); for(int i=0;i<n;i++) for(int j=0;j<=i;j++)cin>>dp[i][j]; for(int i=n-2;i>=0;i--) for(int j=0;j<=i;j++){ dp[i][j]+=max(dp[i+1][j],dp[i+1][j+1]); } cout<<dp[0][0]<<endl; } return 0; }
标签:dp
原文地址:http://blog.csdn.net/mummyding/article/details/43538929