Bone Collector(0-1背包模板)
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Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 51860 Accepted Submission(s): 21837
Problem Description
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 231).
Sample Input
15 101 2 3 4 55 4 3 2 1
Sample Output
14
0-1背包是最基础的背包问题,特点是:每种物品仅有一件,可以选择放或不放。
用子问题定义状态:即f[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。则其状态转移方程便是:
f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}
这个方程非常重要,基本上所有跟背包相关的问题的方程都是由它衍生出来的。所以有必要将它详细解释一下:
“将前i件物品放入容量为v的背包中”这个子问题,若只考虑第i件物品的策略(放或不放),那么就可以转化为一个只牵扯前i-1件物品的问题。
在前i件物品放进容量v的背包时,它有两种情况:
f[i-1][v]:如果不放第i件物品,那么问题就转化为“前i-1件物品放入容量为v的背包中”,价值为f[i-1][v];
f[i-1][v-c[i]]+w[i]:如果放第i件物品,那么问题就转化为“前i-1件物品放入剩下的容量为v-c[i]的背包中”,此时能获得的最大价值就是f[i-1][v-c[i]]再加上通过放入第i件物品获得的价值w[i]。
最后比较第一种与第二种所得价值的大小,哪种相对大,f[i][v]的值就是哪种(这里是重点,理解!)。
用子问题定义状态:即f[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。则其状态转移方程便是:
f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}
这个方程非常重要,基本上所有跟背包相关的问题的方程都是由它衍生出来的。所以有必要将它详细解释一下:
“将前i件物品放入容量为v的背包中”这个子问题,若只考虑第i件物品的策略(放或不放),那么就可以转化为一个只牵扯前i-1件物品的问题。
在前i件物品放进容量v的背包时,它有两种情况:
f[i-1][v]:如果不放第i件物品,那么问题就转化为“前i-1件物品放入容量为v的背包中”,价值为f[i-1][v];
f[i-1][v-c[i]]+w[i]:如果放第i件物品,那么问题就转化为“前i-1件物品放入剩下的容量为v-c[i]的背包中”,此时能获得的最大价值就是f[i-1][v-c[i]]再加上通过放入第i件物品获得的价值w[i]。
最后比较第一种与第二种所得价值的大小,哪种相对大,f[i][v]的值就是哪种(这里是重点,理解!)。
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int w[1010]={0};int c[1010]={0};int val[1010];int main(){int t,i,j;int N,V;scanf("%d",&t);while(t--){memset(val,0,sizeof(val));scanf("%d%d",&N,&V);for(i=1;i<=N;i++){scanf("%d",&w[i]);}for(i=1;i<=N;i++){scanf("%d",&c[i]);}for(i=1;i<=N;i++){for(j=V;j>=c[i];j--){val[j]=max(val[j],val[j-c[i]]+w[i]);//这里可能 刚学的时候不理解,自己模拟一下就懂了; }}printf("%d\n",val[V]);}return 0;}
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