hdu2602 01背包 题解
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习DP,甚是迷惑,审其题,虽解其意,但无从下口,故弃而学背包,今日,重拾之,遂发现其乃美味也。故知背包乃DP之基也~
嘻嘻~~小小嘚瑟一下
我是刚开始接触算法,就被诸位大神弄去学了DP,但是一知半解,遇到这道题的时候,怎么也想不出该怎么做。在我的师兄们(比我高一级的小菜,嘘,别让他听到了)的指引下,看了背包,今天再来看这道题~~~~大神 你在逗我玩吗~~~~~
建议刚开始学dp的人 , 一定要去看看背包九讲,很精彩~~~
废话少数,切入正题
E - Bone Collector
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?
Input
The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.
Output
One integer per line representing the maximum of the total value (this number will be less than 2 31).
Sample Input
15 101 2 3 4 55 4 3 2 1
Sample Output
14
题意分析:一个骨头搜集爱好者,收集各类各样的骨头,这个骨头爱好者有一个大袋子,容积是一定的,现给出一定的骨头,每个骨头的容积和价值是固定的,现在我们往袋子中装入骨头,求袋子容积所能承受的条件下的可装的最大价值。
思路分析:这是个典型的01背包,状态转移方程模板如下
for(int i = 1; i <= N; i++)//N骨头的总数{ for(int j = V; j <= v[i]; j--)//本题中,V指的是袋子的容积,v[i]指第i个骨头的体积,d[i]指第i个骨头的价值 { dp[j]=max(dp[j],dp[j-v[i]]+d[i]); }}
现给出AC代码
#include <iostream>#include<cstdio>#include<string.h>using namespace std;const int MAX = 1000+10;int dp[MAX];int d[MAX];int v[MAX];int T,N,V;int main(){ scanf("%d",&T); while(T--) { memset(dp,0,sizeof(dp)); scanf("%d%d",&N,&V); for(int i = 1; i <= N; i++) { scanf("%d",&d[i]); } for(int j = 1; j <= N; j++) { scanf("%d",&v[j]); } for(int i = 1; i <= N; i++) { for(int j = V; j >= v[i];j--) { dp[j]=max(dp[j],dp[j-v[i]]+d[i]); } } printf("%d\n",dp[V]); } return 0;}
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