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题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=2546
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14019 Accepted Submission(s): 4883
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <string> 5 #include <algorithm> 6 using namespace std; 7 int n, m; 8 int p[1010]; 9 int dp[1010][1010]; 10 int main(){ 11 while(scanf("%d", &n) && n){ 12 for(int i = 1; i <= n; i++) scanf("%d", &p[i]); 13 sort(p+1,p+1+n); 14 //要使剩余体积最小。 15 memset(dp, 0, sizeof(dp)); 16 scanf("%d", &m); 17 if(m >= 5){ 18 for(int i = 1; i <= n-1; i++){ 19 for(int j = 0; j <= m-5; j++){ 20 dp[i][j] = dp[i-1][j]; 21 if(j >= p[i]){ 22 dp[i][j] = max(dp[i][j], dp[i-1][j-p[i]] + p[i]); 23 } 24 } 25 } 26 cout<<m-dp[n-1][m-5]-p[n]<<endl; 27 } 28 else cout<<m<<endl; 29 } 30 31 return 0; 32 }
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原文地址:http://www.cnblogs.com/titicia/p/4356432.html