标签：hdu2602
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28365 Accepted Submission(s): 11562
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 ?
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 231).
Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
01背包入门题。
#include <stdio.h>
#include <string.h>
#define maxn 1002
int dp[maxn], w[maxn], v[maxn];
int main()
{
int t, n, val, i, j;
scanf("%d", &t);
while(t--){
scanf("%d%d", &n, &val);
for(i = 1; i <= n; ++i) scanf("%d", v + i);
for(i = 1; i <= n; ++i) scanf("%d", w + i);
memset(dp, 0, sizeof(dp));
for(i = 1; i <= n; ++i){
for(j = val; j >= w[i]; --j){
if(dp[j] < dp[j-w[i]] + v[i])
dp[j] = dp[j-w[i]] + v[i];
}
}
printf("%d\n", dp[val]);
}
return 0;
}
HDU2602 Bone Collector 【01背包】
标签：hdu2602
原文地址：http://blog.csdn.net/chang_mu/article/details/37973105
