hdu 4296 Buildings 贪心解题报告

楼层堆叠优化问题

最新推荐文章于 2019-02-08 20:13:24 发布

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最新推荐文章于 2019-02-08 20:13:24 发布

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探讨如何通过优化楼层堆叠顺序来最小化建筑的整体潜在损害值，介绍了一种基于权重和强度属性的贪心算法实现。

Buildings

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3457 Accepted Submission(s): 1291

Problem Description

　　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 w_i and strength s_i. When floors are stacked up, each floor has PDV(Potential Damage Value) equal to (Σw_j)-s_i, where (Σw_j) 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.

Input

　　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 w_i and s_i (0 <= w_i, s_i <= 100000) separated by single spaces.
　　Please process until EOF (End Of File).

Output

　　For each test case, your program should output a single integer in a single line - the minimal PDV of the whole building.
　　If no floor would be damaged in a optimal configuration (that is, minimal PDV is non-positive) you should output 0.

Sample Input

Sample Output

1
0
2

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。

思路：

设i在j上面
1)a=sum-si;b=sum+wi-sj;
交换位置得
2)a'=sum+wj-si;b'=sum-sj;

要让1优于2，则

如果1中，b是代表大的，则b因为肯定大于b',所以b要小于a',即wi-sj<wj-si
如果1中，a是代表大的，则a因为肯定小于a',符合要求

所以满足 wi+si < wj+sj ，即按这个来贪心，即可。

代码：

#include <iostream>
#include <cstring>
#include <string>
#include <queue>
#include <cstdio>
#include <algorithm>
using namespace std;

#define MAXN 100000 + 10

struct node
{
    int w, s;
}w[MAXN];

int cmp(node a, node b)
{
    return a.w + a.s < b.w + b.s;
}

int main()
{
    int n;
    while(scanf("%d", &n) != EOF)
    {
        for(int i = 0; i < n; i++)
            scanf("%d%d", &w[i].w, &w[i].s);
        sort(w, w + n, cmp);
        long long sum = 0, maxx = 0;
        for(int i = 0; i < n; i++)
        {
            maxx = max(maxx, sum - w[i].s);
            sum += w[i].w;
        }
        printf("%lld\n", maxx);
    }
    return 0;
}