[2602]Bone Collector (HDU)
2014-08-23 20:03
204 查看
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status
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 ?
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.
Output
One integer per line representing the maximum of the total value (this number will be less than 2 31).
Sample Input
Sample Output
01背包
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
int w[1100],v[1100],bg[1100];
int main()
{
int z,n,m,i,j;
scanf("%d",&z);
while(z--)
{
memset(bg,0,sizeof(bg));
scanf("%d %d",&n,&m);
getchar();
for(i=0;i<n;i++)
scanf("%d",&w[i]);
for(i=0;i<n;i++)
scanf("%d",&v[i]);
for(i=0;i<n;i++)
{
for(j=m;j>=v[i];j--)//从后向前
{
if(bg[j]<bg[j-v[i]]+w[i])
bg[j]=bg[j-v[i]]+w[i];
}
}
printf("%d\n",bg[m]);
}
return 0;
}
Submit Status
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 ?
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.
Output
One 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
01背包
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
int w[1100],v[1100],bg[1100];
int main()
{
int z,n,m,i,j;
scanf("%d",&z);
while(z--)
{
memset(bg,0,sizeof(bg));
scanf("%d %d",&n,&m);
getchar();
for(i=0;i<n;i++)
scanf("%d",&w[i]);
for(i=0;i<n;i++)
scanf("%d",&v[i]);
for(i=0;i<n;i++)
{
for(j=m;j>=v[i];j--)//从后向前
{
if(bg[j]<bg[j-v[i]]+w[i])
bg[j]=bg[j-v[i]]+w[i];
}
}
printf("%d\n",bg[m]);
}
return 0;
}
相关文章推荐
- HDU 2602 Bone Collector
- hdu 2602 Bone Collector (01背包 入门)
- HDU - 2602 Bone Collector —— 01背包
- HDU 2602 (Bone Collector )基础 01背包
- HDU 2602 - Bone Collector
- hdu2602 Bone Collector(背包问题)
- HDU 2602 Bone Collector 01背包(裸) .
- HDU 2602 - Bone Collector
- hdu 2602 Bone Collector
- hdu 2602 Bone Collector
- HDU 2602 Bone Collector (01背包)
- !HDU 2602 Bone Collector--DP--(裸01背包)
- hdu 2602 Bone Collector(0/1背包)
- HDU2602 Bone Collector 01背包入门
- HDU 2602 Bone Collector(最简单的0/1背包问题)
- HDU 2602 Bone Collector(0/1背包)
- HDU 2602 Bone Collector 解题报告(dp入门题)
- hdu2602 — Bone Collector
- HDU 2602--Bone Collector【01背包】
- HDU 2602 Bone Collector(01背包问题)