HDU 2124 Repair the Wall(贪心)

最新推荐文章于 2021-03-12 22:03:06 发布

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原文链接：http://www.cnblogs.com/gpsx/p/5187010.html

本文描述了一个关于使用不同尺寸的木块修复墙壁裂缝的算法问题。玩家需要通过锯断或直接使用木块来填补大小为1×L英寸的矩形裂缝，目标是最少使用木块数量完成修复。文章提供了详细的解决方案，包括输入输出样例和实现代码。

Repair the Wall

Problem Description

Long time ago , Kitty lived in a small village. The air was fresh and the scenery was very beautiful. The only thing that troubled her is the typhoon.

When the typhoon came, everything is terrible. It kept blowing and raining for a long time. And what made the situation worse was that all of Kitty's walls were made of wood.

One day, Kitty found that there was a crack in the wall. The shape of the crack is
a rectangle with the size of 1×L (in inch). Luckly Kitty got N blocks and a saw(锯子) from her neighbors.
The shape of the blocks were rectangle too, and the width of all blocks were 1 inch. So, with the help of saw, Kitty could cut down some of the blocks(of course she could use it directly without cutting) and put them in the crack, and the wall may be repaired perfectly, without any gap.

Now, Kitty knew the size of each blocks, and wanted to use as fewer as possible of the blocks to repair the wall, could you help her ?

Input

The problem contains many test cases, please process to the end of file( EOF ).
Each test case contains two lines.
In the first line, there are two integers L(0<L<1000000000) and N(0<=N<600) which
mentioned above.
In the second line, there are N positive integers. The i ^th integer Ai(0<Ai<1000000000 ) means that the i ^th block has the size of 1×Ai (in inch).

Output

For each test case , print an integer which represents the minimal number of blocks are needed.
If Kitty could not repair the wall, just print "impossible" instead.

Sample Input

5 3
3 2 1
5 2
2 1

Sample Output

2
impossible

Answer

简简单单的题意要用这么长一段去说...用一堆木头去补墙，木头可以锯断，如果不够输出impossible，否则输出最少使用的木头。先判断木头总长是否小于需修补的长度，如果小于，输出impossible，否则，从大到小排序，直到总长大于等于需修补的长度。

#include <cstdio>
#include <iostream>
#include <string>
#include <sstream>
#include <cstring>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <map>
#define PI acos(-1.0)
#define ms(a) memset(a,0,sizeof(a))
#define msp memset(mp,0,sizeof(mp))
#define msv memset(vis,0,sizeof(vis))
using namespace std;
//#define LOCAL
int cmp(int a,int b){return a>b;}
int main()
{
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
#endif // LOCAL
    ios::sync_with_stdio(false);
    int wall,n;
    vector<int> v;
    while(cin>>wall>>n)
    {
        v.clear();
        int sum=0;
        for(int i=0;i<n;i++)
        {
            int t;
            cin>>t;
            sum+=t;
            v.push_back(t);
        }
        if(sum<wall)printf("impossible\n");
        else
        {sort(v.begin(),v.end(),cmp);
        int ans=0,cnt=0;
        for(int i=0;i<(int)v.size();i++)
        {
            if(cnt>=wall)break;
            else cnt+=v[i],ans++;
        }
        printf("%d\n",ans);}
    }
    return 0;
}