hdu Bone Collector
Bone Collector |
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
Total Submission(s): 226 Accepted Submission(s): 115 |
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 ?
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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.
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Output
One integer per line representing the maximum of the total value (this number will be less than 231).
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Sample Input
1 5 10 1 2 3 4 5 5 4 3 2 1 |
Sample Output
14 |
Author
Teddy
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Source
HDU 1st “Vegetable-Birds Cup” Programming Open Contest
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Recommend
lcy
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分析: 典型的0-1背包,V不要求用完。
#include<cstdio> #include<algorithm> #include<string> #include<cstring> #include<iostream> using namespace std; typedef struct S { int v,w; }BONE; BONE bone[1010]; int f[1010]; int main() { int T,W,n,i,j,stop; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&W); for(i=0;i<n;++i) scanf("%d",&bone[i].v); for(i=0;i<n;++i) scanf("%d",&bone[i].w); memset(f,0,sizeof(f)); for(i=0;i<n;++i) { for(j=W;j>=bone[i].w;--j) { if(f[j-bone[i].w]+bone[i].v>f[j]) f[j]=f[j-bone[i].w]+bone[i].v; } } printf("%d\n",f[W]); } return 0; }
这条路我们走的太匆忙~拥抱着并不真实的欲望~