HDU 2602 - Bone Collector
2017-03-03 15:16
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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 ?
InputThe 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.
OutputOne integer per line representing the maximum of the total value (this number will be less than 2 31).
Sample Input
Sample Output
纯01背包题……
二维数组写法:
减小空间复杂度,滚动一维数组写法:
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 ?
InputThe 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.
OutputOne integer per line representing the maximum of the total value (this number will be less than 2 31).
Sample Input
1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output
14
纯01背包题……
二维数组写法:
#include<cstdio> #include<algorithm> using namespace std; unsigned long max_weight[1003][1003];//max_weight[i][j]:前i个骨头中,容量为j的背包里能放的下的最大重量 struct type{ int v; int w; }bone[1003]; int main() { int n,V; int t;scanf("%d",&t); while(t--){ scanf("%d%d",&n,&V); for(int i=1;i<=n;i++) scanf("%d",&bone[i].w); for(int i=1;i<=n;i++) scanf("%d",&bone[i].v); for(int j=0;j<=V;j++){ if(bone[1].v <= j) max_weight[1][j]=bone[1].w; else max_weight[1][j]=0; } for(int i=2;i<=n;i++){ for(int j=0;j<=V;j++){ if(j<bone[i].v) max_weight[i][j]=max_weight[i-1][j]; else max_weight[i][j]=max( max_weight[i-1][j] , max_weight[i-1][ (j-bone[i].v) ] + bone[i].w ); } } printf("%d\n",max_weight [V]); } }
减小空间复杂度,滚动一维数组写法:
#include<cstdio> #include<algorithm> using namespace std; unsigned long max_weight[1003]; struct type{ int v; int w; }bone[1003]; int main() { int n,V; int t;scanf("%d",&t); while(t--){ scanf("%d%d",&n,&V); for(int i=1;i<=n;i++) scanf("%d",&bone[i].w); for(int i=1;i<=n;i++) scanf("%d",&bone[i].v); for(int j=0;j<=V;j++){ if(bone[1].v <= j) max_weight[j]=bone[1].w; else max_weight[j]=0; } for(int i=2;i<=n;i++){ for(int j=V;j>=0;j--){ if(j<bone[i].v) max_weight[j]=max_weight[j]; else max_weight[j]=max( max_weight[j] , max_weight[ (j-bone[i].v) ] + bone[i].w ); } } printf("%d\n",max_weight[V]); } }
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