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0-1背包的总结
HDU 2602 Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30300 Accepted Submission(s): 12477
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.
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
题意：t组测试数据，每组先输入n, m， 代表物品的种类和背包的大小。 接下来两行，每行n个数，第一行代表价值；第二行代表花费。每件物品只能用一次！
输出背包所能装下的最大价值！
代码（一）：
// 0-1背包模板题#include <stdio.h>#include <string.h>int w[1002], c[1002];int dp[1002][1002];int max(int a, int b){ return a>b?a:b;}int main(){ int t; int i, j; scanf("%d", &t); int n, m; while(t--) { scanf("%d %d", &n, &m); for(i=1; i<=n; i++) { scanf("%d", &w[i] ); } for(i=1; i<=n; i++) { scanf("%d", &c[i] ); } for(i=0; i<=n; i++) dp[i][0]=0; for(j=0; j<=m; j++) dp[0][j]=0; for(i=1; i<=n; i++ ) { for(j=0; j<=m; j++) { dp[i][j] = dp[i-1][j]; if( j>=c[i] ) dp[i][j] = max(dp[i-1][j], dp[i-1][j-c[i]]+w[i] ); } } printf("%d\n", dp[n][m] ); } return 0;}代码（二）（刘汝佳的书上写的）：
// 0-1背包模板题#include <stdio.h>#include <string.h>int w[1002], c[1002];int dp[1002][1002];int max(int a, int b){ return a>b?a:b;}int main(){ int t; int i, j; scanf("%d", &t); int n, m; while(t--) { scanf("%d %d", &n, &m); for(i=1; i<=n; i++) { scanf("%d", &w[i] ); } for(i=1; i<=n; i++) { scanf("%d", &c[i] ); } for(i=0; i<=n; i++) dp[i][0]=0; for(j=0; j<=m; j++) dp[0][j]=0; for(i=1; i<=n; i++ ) { for(j=0; j<=m; j++) { dp[i][j] = (i==1?0:dp[i-1][j] ); if( j>=c[i] ) dp[i][j] = max(dp[i][j], dp[i-1][j-c[i]]+w[i] ); } } printf("%d\n", dp[n][m] ); } return 0;}
0-1背包的总结
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