WA了两次,因为没看清题目:花是不能不插的!
d[i][j] = max ( d[i][j - 1], d[i - 1][j - 1] + a[i][j] );表示前i束花,放在前j个花瓶的最大审美数。
// poj 1157 (IOI 1999 - dp)
// [6/13/2014 wind]
#include <iostream>
#include <cstdio>
using namespace std;
int const MAX_N = 101, INF = 0x3f3f3f3f;
int f, v;
int a[MAX_N][MAX_N];
int d[MAX_N][MAX_N];
void solve2()
{
for ( int i = 1; i <= f; i++ )
{
d[i][i] = d[i-1][i-1] + a[i][i];
for ( int j = i + 1; j <= v - f + i; j++ )
{
d[i][j] = max ( d[i][j - 1], d[i - 1][j - 1] + a[i][j] );
}
}
printf ( "%d\n", d[f][v] );
}
int main()
{
freopen ( "in.txt", "r", stdin );
scanf ( "%d%d", &f, &v );
for ( int i = 1; i <= f; i++ )
{
for ( int j = 1; j <= v; j++ )
{
scanf ( "%d", &a[i][j] );
}
}
solve2();
return 0;
}