BZOJ 1618: [Usaco2008 Nov]Buying Hay 购买干草( dp )

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
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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

Sample Output

9


FJ can buy three packages from the second supplier for a total cost of 9.

HINT

Source

Silver