标签：动态规划 acm algorithm hdu c语言
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31922 Accepted Submission(s): 13138
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
Author
Teddy
Source
相信大家都对DP很熟了...但是我不是很熟.....上次北京现场赛就考了好多DP....差点压制死:-(
先贴个代码纪念纪念：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dp[1005][1005], val[1005], vol[1005];
int main()
{
int n, v, T;
scanf("%d", &T);
while(T--)
{
int i, j;
memset(dp, 0, sizeof(dp));
scanf("%d %d", &n, &v);
for(i=1; i<=n; i++)
{
scanf("%d", &val[i]);
}
for(i=1; i<=n; i++)
{
scanf("%d", &vol[i]);
}
for(i=1; i<=n; i++)
{
for(j=0; j<=v; j++)
{
if(j >= vol[i]) dp[i][j] = max(dp[i-1][j], dp[i-1][j-vol[i]]+val[i]);
else dp[i][j] = dp[i-1][j];
}
}
printf("%d\n", dp[n][v]);
}
return 0;
}
HDU-2602-Bone Collector （最基础DP！！）
标签：动态规划 acm algorithm hdu c语言
原文地址：http://blog.csdn.net/u014355480/article/details/41382165
