hdu-1114 Piggy-Bank---完全背包

Posted 2020-10-31 努力努力再努力x

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1114

题目大意：

给出小猪钱罐的重量和装满钱后的重量，然后是几组数据，每组数据包括每种钱币的价值与重量

要求出重量最少能装满钱罐时的最大价值

思路：

完全背包裸题，dp[j] = min(dp[j], dp[j-w[i]]+v[i])

注意dp数组开的范围，一开始开小了，一直WA

 1 #include<iostream>
 2 #include<vector>
 3 #include<queue>
 4 #include<algorithm>
 5 #include<cstring>
 6 #include<cstdio>
 7 #include<set>
 8 #include<map>
 9 #include<cmath>
10 #include<sstream>
11 using namespace std;
12 typedef pair<int, int> Pair;
13 typedef long long ll;
14 const int INF = 0x3f3f3f3f;
15 int T, n, m, minans;
16 const int maxn = 1e3 + 10;
17 int dir[4][2] = {1,0,0,1,-1,0,0,-1};
18 int w[maxn], v[maxn];
19 int dp[100000];
20 int main()
21 {
22     scanf("%d", &T);
23     while(T--)
24     {
25         scanf("%d%d", &n, &m);
26         m -= n;
27         scanf("%d", &n);
28         for(int i = 0; i < n; i++)scanf("%d%d", &v[i], &w[i]);
29         memset(dp, INF, sizeof(dp));
30         dp[0] = 0;
31         for(int i = 0; i < n; i++)
32         {
33             for(int j = w[i]; j <= m; j++)
34                 dp[j] = min(dp[j], dp[j - w[i]] + v[i]);
35         }
36         if(dp[m] >= 1000000)
37             printf("This is impossible.\n");
38         else printf("The minimum amount of money in the piggy-bank is %d.\n", dp[m]);
39     }
40     return 0;
41 }