HDU 2602 Bone Collector(01背包)
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Bone Collector
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 ?
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.
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
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
Answer
01背包,详见注释。
#include <cstdio> #include <iostream> #include <string> #include <sstream> #include <cstring> #include <stack> #include <queue> #include <algorithm> #include <cmath> #include <map> #define PI acos(-1.0) #define ms(a) memset(a,0,sizeof(a)) #define msp memset(mp,0,sizeof(mp)) #define msv memset(vis,0,sizeof(vis)) using namespace std; //#define LOCAL struct Node { int val; int vol; }v[1000]; int dp[1000][1000]; int main() { #ifdef LOCAL freopen("in.txt", "r", stdin); //freopen("out.txt","w",stdout); #endif // LOCAL ios::sync_with_stdio(false); int N; cin>>N; while(N--) { int n,m; cin>>n>>m; ms(dp); for(int i=1;i<=n;i++) cin>>v[i].val; for(int i=1;i<=n;i++) cin>>v[i].vol; for(int i=1;i<=n;i++)//每件物品 { for(int j=0;j<=m;j++)//分别放入容量为[0,m]的背包(物品重量可以是0) { if(v[i].vol<=j)//如果能放入 dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i].vol]+v[i].val); //1.如果前一个大,就不放 //Q:为什么后面的加上了一个可能会比原来小? //A:dp[i-1][j-v[i].vol]表示前i-1个物品,放入j-v[i].vol的背包, //如果放不下,那么这n-1个物品就没有放进去, //也就是说如果放入了第n个物品,背包里只有这个物品, //就可能比不放这个物品而放前n-1个物品要小. //2.后一个大,放 else dp[i][j]=dp[i-1][j];//如果不能放入 } } printf("%d\n",dp[n][m]);//输出[把第n个物品放入m的背包中]获得的最大价值 } return 0; }
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