A Task Process DP

A Task Process

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 615    Accepted Submission(s): 299

Problem Description

There are two kinds of tasks, namely A and B. There are N workers and the i-th worker would like to finish one task A in ai minutes, one task B in bi minutes. Now you have X task A and Y task B, you want to assign each worker some tasks and finish all the tasks as soon as possible. You should note that the workers are working simultaneously.

 

 

Input

In the first line there is an integer T(T<=50), indicates the number of test cases.

In each case, the first line contains three integers N(1<=N<=50), X,Y(1<=X,Y<=200). Then there are N lines, each line contain two integers ai, bi (1<=ai, bi <=1000).

 

 

Output

For each test case, output “Case d: “ at first line where d is the case number counted from one, then output the shortest time to finish all the tasks.

 

 

Sample Input

   
   
   
   

    
    
    
    3
2 2 2
1 10
10 1
2 2 2
1 1
10 10

3 3 3
2 7
5 5
7 2
   
   
   
   

 

 

Sample Output

   
   
   
   

    
    
    
    Case 1: 2
Case 2: 4
Case 3: 6
   
   
   
   

 

 

Author

wzc1989

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（1）——Host by FZU

 

 

Recommend

zhouzeyong

悲剧！比赛时候一点思路也没有！

详细思路见代码注释~

#include<iostream> using namespace std; #define M 60 #define N 210 #define INF 100000000 int cas,tim,n,x,y; int a[60],b[60],f[M][N]; bool Can(int t) { int i,j,k; for (i=0;i<=n;i++) for (j=0;j<=x;j++) f[i][j]=-INF; f[0][0]=0; for (i=1;i<=n;i++) for (j=0;j<=x;j++) for (k=0;k<=j;k++) if ((j-k)*a[i]<=t) f[i][j]=max(f[i][j],f[i-1][k]+(t-(j-k)*a[i])/b[i]); return f[n][x]>=y; } /* f[i,j]=max{f[i-1,k]+(t-(j-k)*a[i])/b[i]}; f[i,j] presents 前i个人完成j个A任务后可以完成的B任务的数量 f[i,j]=max{前i-1个人完成k个A任务时所能完成的B任务个数+剩下的时间还能完成多少B任务} */ int main() { scanf("%d",&cas); for(int tim=0;tim<cas;) { scanf("%d%d%d",&n,&x,&y); for(int i=1;i<=n;i++) scanf("%d%d",&a[i],&b[i]); int low=0,high=a[1]*x+b[1]*y; int mid; while(low<high) { mid=(low+high)>>1; if(Can(mid)) high=mid; else low=mid+1; } printf("Case %d: %d/n",++tim,low); } return 0; }