Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 5823 Accepted: 3802
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:
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
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.
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 5
The highest score is achievable by traversing the cows as shown above.
Source
USACO 2005 December Bronze
//
#include < iostream >
#include < algorithm >
using namespace std;
int main( int argc, char * argv[])
{
int N;
scanf( " %d " , & N);
int balls[ 350 ][ 350 ];
for ( int i = 0 ; i < N; ++ i)
for ( int j = 0 ; j <= i; ++ j) scanf( " %d " , & balls[i][j]);
int DP[ 350 ];
memset(DP, 0 , sizeof (DP));
DP[ 0 ] = balls[ 0 ][ 0 ];
for ( int i = 1 ; i < N; ++ i)
for ( int j = i; j >= 0 ; -- j)
{
if (j == i) DP[j] = DP[j - 1 ] + balls[i][j];
else if (j == 0 ) DP[j] = DP[j] + balls[i][j];
else
DP[j] = max(DP[j - 1 ],DP[j]) + balls[i][j];
};
cout << * max_element( & DP[ 0 ], & DP[N]) << endl;
return 0 ;
}
转载于:https://www.cnblogs.com/asuran/archive/2009/10/13/1582443.html