Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 51860    Accepted Submission(s): 21837 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

题：图不重要，看字

就是某人有体积为v的袋子，要捡骨头，给出n块骨头的价值与体积，问最多拿到多少价值的骨头---恶趣味

01背包模型

先给出二维的吧，二维时/空间都不能压缩

#include<cstdio> #include<cstring> #include<algorithm> #define ll long long #define inf 0x3f3f3f3f #define mes(a, b) memset(a, b, sizeof(a)) #define maxn 1010 using namespace std; int ans, m, n, v; int val[maxn], vol[maxn], dp[maxn][maxn]; int main(){ int t; scanf("%d", &t); while(t--){ mes(val, 0); mes(vol, 0); mes(dp, 0); scanf("%d%d", &n, &v); for(int i = 1; i <= n; i++){ scanf("%d", &val[i]); } for(int i = 1; i <= n; i++){ scanf("%d", &vol[i]); } for(int i = 1; i <= n; i++){ for(int j = 0; j <= v; j++){ if(j >= vol[i]){ dp[i][j] = max(dp[i-1][j], dp[i-1][j-vol[i]]+val[i]); } else{ dp[i][j] = dp[i-1][j]; } } } printf("%d\n", dp[n][v]); } return 0; } 一维的，略微压缩时间与空间

#include<cstdio> #include<cstring> #include<algorithm> #define ll long long #define inf 0x3f3f3f3f #define mes(a, b) memset(a, b, sizeof(a)) #define maxn 1010 using namespace std; int ans, m, n, v; int val[maxn], vol[maxn], dp[maxn]; int main(){ int t; scanf("%d", &t); while(t--){ mes(val, 0); mes(vol, 0); mes(dp, 0); scanf("%d%d", &n, &v); for(int i = 1; i <= n; i++){ scanf("%d", &val[i]); } for(int i = 1; i <= n; i++){ scanf("%d", &vol[i]); } for(int i = 1; i <= n; i++){ for(int j = v; j >= 0; j--){ if(j >= vol[i]){ dp[j] = max(dp[j], dp[j-vol[i]]+val[i]); continue; } break; } } printf("%d\n", dp[v]); } return 0; }