无限背包dp..
因为题目中说至少到 H 磅 , 我就直接把 H * 2 了..
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#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#define rep( i , n ) for( int i = 0 ; i < n ; ++i )
#define clr( x , c ) memset( x , c , sizeof( x ) )
using namespace std;
const int maxn = int( 1e5 ) + 5;
const int inf = 0x3f3f3f3f;
int d[ maxn ];
int main() {
// freopen( "test.in" , "r" , stdin );
int n , h;
cin >> n >> h;
h *= 2;
clr( d , inf );
d[ 0 ] = 0;
rep( i , n ) {
int w , v;
scanf( "%d%d" , &w , &v );
for( int i = w ; i <= h ; i++ )
d[ i ] = min( d[ i ] , d[ i - w ] + v );
}
int ans = inf;
for( int i = h / 2 ; i <= h ; i++ )
ans = min( ans , d[ i ] );
cout << ans << "\n";
return 0;
}
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1618: [Usaco2008 Nov]Buying Hay 购买干草
Time Limit: 5 Sec Memory Limit: 64 MB Submit: 749 Solved: 379 [ ][ ][ ]Description
约翰的干草库存已经告罄,他打算为奶牛们采购日(1≤日≤50000)磅干草.
他知道N(1≤N≤100)个干草公司,现在用1到N给它们编号.第i个公司卖的干草包重量为Pi(1≤Pi≤5000)磅,需要的开销为Ci(l≤Ci≤5000)美元.每个干草公司的货源都十分充足,可以卖出无限多的干草包. 帮助约翰找到最小的开销来满足需要,即采购到至少H磅干草.
Input
第1行输入N和日,之后N行每行输入一个Pi和Ci.
Output
最小的开销.
Sample Input
2 15
3 2
5 3
3 2
5 3
Sample Output
9
FJ can buy three packages from the second supplier for a total cost of 9.
FJ can buy three packages from the second supplier for a total cost of 9.
HINT
Source