HDU 2602 Bone Collector(01背包)

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背包问题，这种背包特点是：每种物品仅有一件，可以选择放或不放。
用子问题定义状态：即dp[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。
则其状态转移方程便是：
dp[i][v]=max{dp[i-1][v],dp[i-1][v-cost[i]]+value[i]}

注意体积为零的情况，如：
1
5 0
2 4 1 5 1
0 0 1 0 0
结果为12

时间复杂度：O(n*v)

//01背包问题：//动态规划//状态转移方程:F[i，Vnow]=max(F[i-1,Vnow],F[i-1,Vow-Vi]+wealthi);#include <cstdio>#include <cstring>#include <iostream>using namespace std;struct bone{int value;int volume;};int dp[1050][1050];int main(){int test;int n,vol;struct bone pack[1050];cin>>test;while(test--){cin>>n>>vol;memset(dp,0,sizeof(dp));for(int i=1;i<=n;i++)cin>>pack[i].value;for(int i=1;i<=n;i++)cin>>pack[i].volume;for(int i=1;i<=n;i++)for(int j=0;j<=vol;j++){if(j-pack[i].volume>=0&&dp[i-1][j]<dp[i-1][j-pack[i].volume]+pack[i].value)dp[i][j]=dp[i-1][j-pack[i].volume]+pack[i].value;elsedp[i][j]=dp[i-1][j];}cout<<dp[n][vol]<<'\n';}return 0;}

该题的第二种解法就是对背包的优化解法，当然只能对空间就行优化，时间是不能优化的。

先考虑上面讲的基本思路如何实现，肯定是有一个主循环i=1..N，每次算出来二维数组dp[i][0..V]的所有值。
那么，如果只用一个数组dp[0..V]，能不能保证第i次循环结束后dp[v]中表示的就是我们定义的状态dp[i][v]呢？

dp[i][v]是由dp[i-1][v]和dp[i-1][v-c[i]]两个子问题递推而来，能否保证在推dp[i][v]时（也即在第i次主循环中推dp[v]时）能够得到dp[i-1][v]和dp[i-1][v-c[i]]的值呢？事实上，这要求在每次主循环中我们以v=V..0的顺序推dp[v]，这样才能保证推dp[v]时dp[v-c[i]]保存的是状态dp[i-1][v-c[i]]的值。伪代码如下：

for i=1..N

    for v=V..0

        dp[v]=max{dp[v],dp[v-c[i]]+w[i]};

注意：这种解法只能由V--0，不能反过来，如果反过来就会造成物品重复放置！

#include<iostream>using namespace std;#define Size 1111int va[Size],vo[Size];int dp[Size];int Max(int x,int y){    return x>y?x:y;}int main(){    int t,n,v;    int i,j;    cin>>t;    while(t--)    {        cin>>n>>v;        for(i=1;i<=n;i++)            cin>>va[i];        for(i=1;i<=n;i++)            cin>>vo[i];        memset(dp,0,sizeof(dp));        for(i=1;i<=n;i++)        {            for(j=v;j>=vo[i];j--)            {                dp[j]=Max(dp[j],dp[j-vo[i]]+va[i]);             }        }        cout<<dp[v]<<endl;    }    return 0;}

以上迭代实现，多为自底向上的

01背包递归实现(自顶向下)：

int v, n;int ary[MAXN][2]; // 0 体积  1 价值int result[MAXN][MAXV];int dp(int n, int v) {    if(n <= 0 || v <= 0) return 0;    else if(result[n][v]) return result[n][v];    if(v >= ary[n][0]) return result[n][v] = max(dp(n - 1, v), dp(n - 1, v - ary[n][0]) + ary[n][1]);    else return result[n][v] = dp(n - 1, v);}