HDU 2602 Bone Collector(背包模板)
2016-08-15 11:53
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Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 51885 Accepted Submission(s): 21849
[align=left]Problem Description[/align]
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 ?
[align=left]Input[/align]
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.
[align=left]Output[/align]
One integer per line representing the maximum of the total value (this number will be less than 231).
[align=left]Sample Input[/align]
1
5 10
1 2 3 4 5
5 4 3 2 1
[align=left]Sample Output[/align]
14
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]}
代码:
<pre name="code" class="html">#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int V[1010]; int value[1010]; int dp[1010][1010]; int main() { int t,n,v; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&v); for(int i=1;i<=n;i++) { scanf("%d",&value[i]); } for(int i=1;i<=n;i++) { scanf("%d",&V[i]); } memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) { for(int j=0;j<=v;j++) { dp[i][j]=dp[i-1][j];//每次都要计算一下不放第i个物品时的价值; if(j>=V[i])//若能放第i个物品; { dp[i][j]=max(dp[i][j],dp[i-1][j-V[i]]+value[i]); } } } printf("%d\n",dp [v]); } return 0; }
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