动态规划1017
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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 …<br>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 ?<br><center><img src=../../../data/images/C154-1003-1.jpg> </center><br>
Input
The first line contain a integer T , the number of cases.<br>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<sup>31</sup>).
Sample Input
15 101 2 3 4 55 4 3 2 1
Sample Output
14
代码:
#include<stdio.h>
#include<string.h>
int f[2002];
int max(int x,int y){
if(x<y)
x=y;
return x;
}
int main()
{
int t,n,V,v[1000],w[1000],i,j;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&V);
for(i=1;i<=n;i++)
scanf("%d",&w[i]);
for(i=1;i<=n;i++)
scanf("%d",&v[i]);
memset(f,0,sizeof(f));
for(i=1;i<=n;i++)
for(j=V;j>=v[i];j--)
f[j]=max(f[j],f[j-v[i]]+w[i]);
printf("%d\n",f[V]);
}
return 0;
}
这是一个01背包问题,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。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];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]
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