TOJ 1635

题目标题:

Space Elevator

题目连接:

http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=1635

题目类型:

动态规划 - 多重背包

数据结构:

struct LMIC_BLOCK{int h, a, c;bool operator < ( const LMIC_BLOCK & K ) const {return a < K.a;}}

思路分析:

多重背包问题,

可以将每种砖块按照各自最大可以达到的高度进行排序,

所限制该种砖块的条件是 该砖块的数量和可以达到的高度

因为每种砖块的数量是有限的

所以是多重背包问题

解答的时候

讲多重背包拆分成01背包即可

证明:

略

源代码:

#include <iostream>#include <stdio.h>using namespace std;struct LMIC_BLOCK{int h, a, c;bool operator < ( const LMIC_BLOCK & K ) const {return a < K.a;}} blk[40005];int dp[40005];void _01pack( int p_k, LMIC_BLOCK p_pack ){int i;p_pack.h *= p_k;for( i = p_pack.a; i >= p_pack.h ; i -- ){dp[i] = max( dp[i], dp[i - p_pack.h] + p_pack.h );}}void _compack( LMIC_BLOCK p_pack ){int i;for( i = p_pack.h; i <= p_pack.a; i ++ ){dp[i] = max( dp[i], dp[i - p_pack.h] + p_pack.h );}}void _multpack( LMIC_BLOCK p_pack ){if( p_pack.h * p_pack.c >= p_pack.a ){_compack( p_pack );return ;}int k = 1;while( k < p_pack.c ){_01pack( k, p_pack );p_pack.c -= k;k = k * 2;}_01pack( p_pack.c, p_pack );}int main(){int i, k;scanf( "%d", &k );{memset( dp, 0, sizeof( dp ) );for( i = 0; i < k; i ++ ){scanf( "%d%d%d", &blk[i].h, &blk[i].a, &blk[i].c );}sort( blk, blk + k );for( i = 0; i < k; i ++ ){_multpack( blk[i] );}int max = -1;for( i = 0; i <= 40004; i ++ ){if( max < dp[i] ){max = dp[i];}}printf( "%d\n", max );}return 0;}/*12 10 1222 12 123 33 1*/