标签：des style blog color os io strong for
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 13481 | Accepted: 8909 |
Description
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. Input
Output
Sample Input
5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5
Sample Output
30
Hint
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.#include<stdio.h> #include<string.h> #include<stdlib.h> #include<algorithm> #define N 351 using namespace std; int main() { int n,i,j,dp[N][N],map[N][N]; scanf("%d",&n); for(i=1;i<=n;i++) for(j=1;j<=i;j++) { scanf("%d",&map[i][j]); } for(i=1;i<=n;i++) dp[n][i]=map[n][i];//最后一行，用新数组存好。运算时新数组其他为原始的0，计算更加好 for(i=n-1;i>=1;i--) for(j=1;j<=i;j++) dp[i][j]=map[i][j]+max(dp[i+1][j],dp[i+1][j+1]); //用到max()的时候，要么自己写一个max比较函数，要么用函数名 algorithm和using namespace std一起 printf("%d",dp[1][1]); return 0; }
标签：des style blog color os io strong for
原文地址：http://www.cnblogs.com/lipching/p/3910054.html
