Bone collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 51972    Accepted Submission(s): 21897 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 2 ³¹).   Sample Input 1 5 10 1 2 3 4 5 5 4 3 2 1   Sample Output 14   Author Teddy  

/* 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 */ #include<stdio.h> #include<string.h> #define max(a,b) a>b?a:b int dp[1000][1000]; int val[1000]; int wig[1000]; int main() { int t; int m,n; int i,j; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(i=1;i<=n;i++) scanf("%d",&val[i]); for(i=1;i<=n;i++) scanf("%d",&wig[i]); memset(dp,0,sizeof(dp));//初始化操作 for(i=1;i<=n;i++) for(j=0;j<=m;j++) { if(j>=wig[i])//表示第i个物品将放入大小为j的背包中 dp[i][j]=max(dp[i-1][j],dp[i-1][j-wig[i]]+val[i]);//第i个物品放入后，那么前i-1个物品可能会放入也可能因为剩余空间不够无法放入 else dp[i][j]=dp[i-1][j];//第i个物品无法放入 } printf("%d\n",dp[n][m]); } return 0; } <pre name="code" class="cpp">/* 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<stdio.h> #include<string.h> #define max(a,b) a>b?a:b int dp[1000]; int val[1000]; int wig[1000]; int main() { int t; int m,n; int i,j; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(i=1;i<=n;i++) scanf("%d",&val[i]); for(i=1;i<=n;i++) scanf("%d",&wig[i]); memset(dp,0,sizeof(dp));//初始化操作 for(i=1;i<=n;i++) for(j=m;j>=0;j--) { if(j>=wig[i])//表示第i个物品将放入大小为j的背包中 dp[j]=max(dp[j],dp[j-wig[i]]+val[i]);//第i个物品放入后，那么前i-1个物品可能会放入也可能因为剩余空间不够无法放入 else dp[j]=dp[j];//第i个物品无法放入 } printf("%d\n",dp[m]); } return 0; }