UVA 11389(贪心问题)

UVA 11389

Time Limit:1000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

 

Description

II U C   ONLINE   C ON TEST   2 008
The Bus Driver Problem
Input: standard input Output: standard output
In a city there are n bus drivers. Also there are n morning bus routes & n afternoon bus routes with various lengths. Each driver is assigned one morning route & one evening route. For any driver, if his total route length for a day exceeds d, he has to be paid overtime for every hour after the first d hours at a flat r taka / hour. Your task is to assign one morning route & one evening route to each bus driver so that the total overtime amount that the authority has to pay is minimized.
Input
The first line of each test case has three integers n, d and r, as described above. In the second line, there are n space separated integers which are the lengths of the morning routes given in meters. Similarly the third line has n space separated integers denoting the evening route lengths. The lengths are positive integers less than or equal to 10000. The end of input is denoted by a case with three 0 s.
Output
For each test case, print the minimum possible overtime amount that the authority must pay.
Constraints
-           1 ≤ n ≤ 100 -           1 ≤ d ≤ 10000 -           1 ≤ r ≤ 5
Sample Input	Output for Sample Input
2 20 5 10 15 10 15 2 20 5 10 10 10 10 0 0 0	50 0

 题解：本题是解决实际问题，有n个上午的任务和下午的任务，分配给司机，如果工作总时间超过d，超过的部分要给加班费；

            现在让你安排任务，问最小的加班分花费。

分析：贪心问题。将两个任务分别按递增和递减序排序，每个对应边号的一组即可。

            设序列中的两组配对的元素为m1，a1，m2，a2 且 m1≤m2，a1≤a2；

            则配对方式<m1，a2>，<m2，a1>优于<m1，a1>，<m2，a2>

代码如下：

#include<iostream>
#include<algorithm>
using namespace std;

int a[105],b[105];

int main()
{

    int i,x,y,z;
    while(cin>>x>>y>>z&&x&&y&&z)
    {
        int s=0;
        for(i=1; i<=x; i++)
        {
            cin>>a[i];
        }
        for(int i=1; i<=x; i++)
        {
            cin>>b[i];
        }
        sort(a,a+x);
        sort(b,b+x);
        for(i=1; i<=x; i++)
        {
            if(a[i]+b[x+1-i]>y)
            {
                s=(a[i]+b[x-i+1]-y)*z;
                s+=s;
            }
        }
        cout<<s<<endl;
    }
    return 0;
}

 咳咳，其实上面是错的。。虽然可以输出正确结果，但是提交不会AC。

以下是正确的代码，请注意改动的地方！！

#include<iostream>
#include<algorithm>
using namespace std;
int a[105],b[105];
int main()
{

    int i,j,k,x,y,z;
    while(cin>>x>>y>>z&&x&&y&&z)
    {
        int s=0;
        for(i=1; i<=x; i++)
        {
            cin>>a[i];
        }
        for(int k=1; k<=x; k++)
        {
            cin>>b[k];
        }
        sort(a,a+x);
        sort(b,b+x);
        for(int j=1; j<=x; j++)
        {
            if(a[j]+b[x+1-j]>y)
            {
                s=(a[j]+b[x-j+1]-y)*z;
                s+=s;
            }
        }
        cout<<s<<endl;
    }
    return 0;
}

可以仔细想一想为什么会出现这样的情况