Bone Collector

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 51847????Accepted Submission(s): 21829

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

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Author Teddy

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Source HDU 1st “Vegetable-Birds Cup” Programming Open Contest

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Recommend lcy???|???We have carefully selected several similar problems for you:??1203?2159?2955?1171?2191? 狀態轉移方程：dp[i][j] = max{dp[i+1][j], dp[i+1][j-vol[i]]+val[i]};其中i表示第i個物品，j表示背包當前的容量，而數組中所存放的元素為在當前狀態下的最大價值。通過下面的每個狀態的最大值向上推。因此，最終的最大狀態應該存放在dp[0][v]中。
#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> using namespace std; const int MAXN = 1000 + 10; const int INF = 0x3f3f3f3f; int n, v, val[MAXN], vol[MAXN], dp[MAXN][MAXN]; int main() {int t; scanf("%d", &t);while (t--) {scanf("%d %d", &n, &v);for (int i = 0; i < n; i++) {scanf("%d", &val[i]);}for (int i = 0; i < n; i++) {scanf("%d", &vol[i]);}memset(dp, 0, sizeof(dp));for (int i = n - 1; i >= 0; i--) {for (int j = 0; j <= v; j++) {if (j < vol[i]) dp[i][j] = dp[i + 1][j];else dp[i][j] = max(dp[i+1][j], dp[i+1][j-vol[i]]+val[i]);}}printf("%d\n", dp[0][v]);}return 0; }

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轉載于:https://www.cnblogs.com/cniwoq/p/6770830.html

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