hdu4415 贪心

http://acm.hdu.edu.cn/showproblem.php?pid=4415

Problem Description

Ezio Auditore is a great master as an assassin. Now he has prowled in the enemies’ base successfully. He finds that the only weapon he can use is his cuff sword and the sword has durability m. There are n enemies he wants to kill and killing each enemy needs Ai durability. Every time Ezio kills an enemy he can use the enemy’s sword to kill any other Bi enemies without wasting his cuff sword’s durability. Then the enemy’s sword will break. As a master, Ezio always want to do things perfectly. He decides to kill as many enemies as he can using the minimum durability cost.

 

Input

The first line contains an integer T, the number of test cases.
For each test case:
The first line contains two integers, above mentioned n and m (1<=n<=10^5, 1<=m<=10^9).
Next n lines, each line contains two integers Ai, Bi. (0<=Ai<=10^9, 0<=Bi<=10).

 

Output

For each case, output "Case X: " (X is the case number starting from 1) followed by the number of the enemies Ezio can kill and the minimum durability cost.

 

Sample Input

23 54 15 17 72 12 24 0

 

Sample Output

Case 1: 3 4Case 2: 0 0

 

/**hdu4415 贪心我考虑少了，就不说什么了……http://m.blog.csdn.net/blog/HELLO_THERE/8016365*/#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>using namespace std;const int maxn=100005;int a[maxn],b[maxn],temp[maxn];int n,m;int main(){    int T,tt=0;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        int k1=0,k2=0,sword=0;        for(int i=0;i<n;i++)        {            int x,y;            scanf("%d%d",&x,&y);            y==0?(b[k2++]=x):(a[k1++]=x);            sword+=y;        }        sort(a,a+k1);        sort(b,b+k2);        int cost1=0,num1=0;        for(int i=0;i<k2;i++)        {            if(m-cost1>=b[i])            {                cost1+=b[i];                num1++;            }            else                break;        }        int num2=0,cost2=0;        if(m>=a[0])        {           if(sword>=n-1)           {                if(num1<n)                {                    num1=n;                    cost1=a[0];                }                else if(num1==n&&cost1>a[0])                {                    cost1=a[0];                }                printf("Case %d: %d %d\n",++tt,num1,cost1);                continue;           }           int k=n-1-sword;           int x=1,y=0,i=0;           a[k1]=b[k2]=1000000000;           while(i<k)           {               if(a[x]<b[y])                   temp[i]=a[x++];               else                   temp[i]=b[y++];               i++;           }           i=0;           num2=1+sword;           cost2=a[0];           while(i<k&&cost2<=m)           {               if(num2==n)                    break;               if(m-cost2>=temp[i])               {                   cost2+=temp[i++];                   num2++;               }               else                   break;           }        }        if(num2>num1)        {            num1=num2;            cost1=cost2;        }        else if(num2==num1&&cost2<cost1)        {            cost1=cost2;        }        printf("Case %d: %d %d\n",++tt,num1,cost1);    }    return 0;}