Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17590 Accepted Submission(s): 6954
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
Sample Output
14
最近比较忙，在看01背包，也不知道为什么，好久不做题了，似乎什么也不会，01背包的一维的优化怎么也看不懂，今天看了一个博客，画了一个表，终于明白了。菜鸟还要努力啊！搞ＡＣＭ这么长时间连个动态规划还处于入门阶段……
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <algorithm> 6 using namespace std; 7 int main(void){ 8 #ifndef ONLINE_JUDGE 9 freopen("2602.in", "r", stdin); 10 #endif 11 const int N = 1000+10; 12 int t, n, V, w[N], c[N], f[N]; 13 scanf("%d", &t); 14 while (t--){ 15 scanf("%d%d", &n, &V); 16 for (int i = 1; i <= n; ++i) scanf("%d", w+i); 17 for (int i = 1; i <= n; ++i) scanf("%d", c+i); 18 memset(f, 0, sizeof(f)); 19 for (int i = 1; i <= n; ++i){ 20 for (int v = V; v >= c[i]; --v){ 21 f[v] = max(f[v], f[v-c[i]] + w[i]); 22 } 23 } 24 printf("%d\n", f[V]); 25 } 26 return 0; 27 }
引用一个博客，用于理解01背包，尤其是那个图表，自己画一遍，就基本理解为什么可以优化成一维数组了！