标签:queue def name pid get pair 16px max http
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 }
标签:queue def name pid get pair 16px max http
原文地址:https://www.cnblogs.com/fzl194/p/8823940.html