poj3176--Cow Bowling(dp:数塔问题)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 14028 | Accepted: 9302 |
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.
Given a triangle with N (1 <= N <= 350) rows, determine the highest possible sum achievable.
Input
Lines 2..N+1: Line i+1 contains i space-separated integers that represent row i of the triangle.
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
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dp[400][400] ;
int main()
{
int n , i , j , max1 ;
while(scanf("%d", &n) !=EOF)
{
max1 = 0 ;
memset(dp,0,sizeof(dp));
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] ) + dp[i][j] ;
for(i = 1 ; i <= n ; i++)
max1 = max(max1,dp[n][i]);
printf("%d\n", max1);
}
return 0;
}
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