POJ 1742 Coins 多重背包用单调队列优化
2014-05-27 22:37
295 查看
多重背包用单调队列优化
Description
People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the
exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch.
You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.
Input
The input contains several test cases. The first line of each test case contains two integers n(1<=n<=100),m(m<=100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1<=Ai<=100000,1<=Ci<=1000). The
last test case is followed by two zeros.
Output
For each test case output the answer on a single line.
Sample Input
Sample Output
多重背包用单调队列优化
Description
People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the
exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch.
You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.
Input
The input contains several test cases. The first line of each test case contains two integers n(1<=n<=100),m(m<=100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1<=Ai<=100000,1<=Ci<=1000). The
last test case is followed by two zeros.
Output
For each test case output the answer on a single line.
Sample Input
3 10 1 2 4 2 1 1 2 5 1 4 2 1 0 0
Sample Output
8 4
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn=100000+10; int c[maxn],w[maxn],use[maxn],f[maxn]; int main() { int n,m; while(scanf("%d%d",&n,&m)&&n&&m) { int ans=0; for(int i=1;i<=n;i++) { scanf("%d",&w[i]); } for(int i=1;i<=n;i++) { scanf("%d",&c[i]); } memset(f,0,sizeof(f)); f[0]=1; for(int i=1;i<=n;i++) { memset(use,0,sizeof(use)); for(int j=w[ 4000 i];j<=m;j++) { if(!f[j]&&f[j-w[i]]&&(use[j-w[i]]+1<=c[i])) { f[j]=1; use[j]=use[j-w[i]]+1; ans++; } } } printf("%d\n",ans); } return 0; }
相关文章推荐
- POJ 1742 coins 多重背包单调队列优化
- POJ 1742:Coins——单调队列优化的多重背包
- poj1742 Coins(多重背包+单调队列优化)
- POJ 1742 Coins(多重背包 + 单调队列优化)
- POJ - 1742 Coins 多重背包+(二进制优化||单调队列优化)
- POJ 1742 Coins( 单调队列优化多重背包)
- POJ 1742 Coins——不要套单调队列优化多重背包的模板
- POJ 1742 Coins 多重背包(单调队列优化)
- POJ 1742 Coins 多重背包单调队列优化
- poj1742 单调队列优化多重背包
- POJ 1742 Coins(多重背包, 单调队列)
- 单调队列优化多重背包(含构造问题<POJ 1742 coin>)
- poj 1742 多重背包 (单调队列优化)
- POJ 3260 The Fewest Coins(完全背包 + 多重背包 + 单调队列优化)
- POJ 1742 Coins 单调队列多重背包
- POJ 1742 Coins 优化后的多重背包
- poj1014多重背包--单调队列优化
- POJ 1742 coins(背包+二进制优化+bitset)
- poj 1742 Coins(dp之多重背包+多次优化)
- poj 2373 单调队列优化背包