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2015
05-23

HDU 4296-Buildings-贪心-[解题报告]HOJ

Buildings

问题描述 :

  Have you ever heard the story of Blue.Mary, the great civil engineer? Unlike Mr. Wolowitz, Dr. Blue.Mary has accomplished many great projects, one of which is the Guanghua Building.
  The public opinion is that Guanghua Building is nothing more than one of hundreds of modern skyscrapers recently built in Shanghai, and sadly, they are all wrong. Blue.Mary the great civil engineer had try a completely new evolutionary building method in project of Guanghua Building. That is, to build all the floors at first, then stack them up forming a complete building.
  Believe it or not, he did it (in secret manner). Now you are face the same problem Blue.Mary once stuck in: Place floors in a good way.
  Each floor has its own weight wi and strength si. When floors are stacked up, each floor has PDV(Potential Damage Value) equal to (Σwj)-si, where (Σwj) stands for sum of weight of all floors above.
  Blue.Mary, the great civil engineer, would like to minimize PDV of the whole building, denoted as the largest PDV of all floors.
  Now, it’s up to you to calculate this value.

输入:

  There’re several test cases.
  In each test case, in the first line is a single integer N (1 <= N <= 105) denoting the number of building’s floors. The following N lines specify the floors. Each of them contains two integers wi and si (0 <= wi, si <= 100000) separated by single spaces.
  Please process until EOF (End Of File).

输出:

  There’re several test cases.
  In each test case, in the first line is a single integer N (1 <= N <= 105) denoting the number of building’s floors. The following N lines specify the floors. Each of them contains two integers wi and si (0 <= wi, si <= 100000) separated by single spaces.
  Please process until EOF (End Of File).

样例输入:

3
10 6
2 3
5 4
2
2 2
2 2
3
10 3
2 5
3 3

样例输出:

1
0
2

题目:http://acm.hdu.edu.cn/showproblem.php?pid=4296

题意:有n个板,每个板有重量和强度两个属性,把板叠在一起,对于每个板有个PDV值,计算方式为这个板上面的板的重量和减去这个板的强度,对于每种叠放方式,取这个叠放方式中所以板中PDV值最大的值为代表值,问所有叠放方式中最小的代表值为多少。

题解:对于相邻放置的两块板,设两块板为i,j他们上面的重量为sum

           1) a=sum-si;b=sum+wi-sj;

           交换两个板的位置

          2)a’=sum+wj-si;b’=sum-sj;

          如果1优于2,求解得有效的条件为wj-si>wi-sj。

          所以按si+wi的和排序贪心即可。

代码:

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
struct point
{
    int w,s;
}node[100005];
int n;
bool cmp(const point &a,const point &b)
{
    return a.w+a.s<b.w+b.s;
}
int main()
{
    for(;~scanf("%d",&n);)
    {
        for(int i=0;i<n;++i)
            scanf("%d%d",&node[i].w,&node[i].s);
        sort(node,node+n,cmp);
        LL summ=0,maxx=0;
        for(int i=0;i<n;++i)
        {
            maxx=max(maxx,summ-node[i].s);
            summ+=node[i].w;
        }
        printf("%I64d\n",maxx);
    }
    return 0;
}

来源:http://blog.csdn.net/acm_ted/article/details/7984935

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参考:http://blog.csdn.net/acm_ted/article/details/7984935