題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=2602

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 54132????Accepted Submission(s): 22670

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 ?

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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.

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Output One integer per line representing the maximum of the total value (this number will be less than 2³¹).

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Sample Input 1 5 10 1 2 3 4 5 5 4 3 2 1

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Sample Output 14 01背包問題，這種背包特點是：每種物品僅有一件，可以選擇放或不放。
用子問題定義狀態：即dp[i][v]表示前i件物品恰放入一個容量為v的背包可以獲得的最大價值。
則其狀態轉移方程便是：
dp[i][v]=max{dp[i-1][v],dp[i-1][v-cost[i]]+value[i]} 1 #include<iostream> 2 using namespace std; 3 int dp[1000][1000]; 4 5 int max(int x,int y) 6 { 7 return x>y?x:y; 8 } 9 10 int main() 11 { 12 int t,n,v,i,j; 13 int va[1000],vo[1000]; 14 cin>>t; 15 while(t--) 16 { 17 cin>>n>>v; 18 for(i=1;i<=n;i++) 19 cin>>va[i]; 20 for(i=1;i<=n;i++) 21 cin>>vo[i]; 22 memset(dp,0,sizeof(dp));//初始化操作 23 for(i=1;i<=n;i++) 24 { 25 for(j=0;j<=v;j++) 26 { 27 if(vo[i]<=j)//表示第i個物品將放入大小為j的背包中 28 dp[i][j]=max(dp[i-1][j],dp[i-1][j-vo[i]]+va[i]);//第i個物品放入后，那么前i-1個物品可能會放入也可能因為剩余空間不夠無法放入 29 else //第i個物品無法放入 30 dp[i][j]=dp[i-1][j]; 31 } 32 } 33 cout<<dp[n][v]<<endl; 34 } 35 return 0; 36 }

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該題的第二種解法就是對背包的優化解法，當然只能對空間就行優化，時間是不能優化的。

先考慮上面講的基本思路如何實現，肯定是有一個主循環i=1..N，每次算出來二維數組dp[i][0..V]的所有值。
那么，如果只用一個數組dp[0..V]，能不能保證第i次循環結束后dp[v]中表示的就是我們定義的狀態dp[i][v]呢？

dp[i][v]是由dp[i-1][v]和dp[i-1][v-c[i]]兩個子問題遞推而來，能否保證在推dp[i][v]時（也即在第i次主循環中推dp[v]時）能夠得到dp[i-1][v]和dp[i-1][v-c[i]]的值呢？事實上，這要求在每次主循環中我們以v=V..0的順序推dp[v]，這樣才能保證推dp[v]時dp[v-c[i]]保存的是狀態dp[i-1][v-c[i]]的值。偽代碼如下：

for i=1..N

??? for v=V..0

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

注意：這種解法只能由V--0，不能反過來，如果反過來就會造成物品重復放置！

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1 #include<iostream> 2 using namespace std; 3 #define Size 1111 4 int va[Size],vo[Size]; 5 int dp[Size]; 6 int Max(int x,int y) 7 { 8 return x>y?x:y; 9 } 10 int main() 11 { 12 int t,n,v; 13 int i,j; 14 cin>>t; 15 while(t--) 16 { 17 cin>>n>>v; 18 for(i=1;i<=n;i++) 19 cin>>va[i]; 20 for(i=1;i<=n;i++) 21 cin>>vo[i]; 22 memset(dp,0,sizeof(dp)); 23 for(i=1;i<=n;i++) 24 { 25 for(j=v;j>=vo[i];j--) 26 { 27 dp[j]=Max(dp[j],dp[j-vo[i]]+va[i]); 28 } 29 } 30 cout<<dp[v]<<endl; 31 } 32 return 0; 33 }

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轉載于:https://www.cnblogs.com/yoke/p/6106079.html

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