zoukankan      html  css  js  c++  java
  • ZOJ 3194 Coverage【贪心】【思维题】

    Coverage

    Time Limit: 1 Second      Memory Limit: 32768 KB

    Given n points (xi, yi) (i = 1, 2 ... n) in a plane, where all xi will be distinct. After connecting the n points with staight lines in order from the leftmost point to the rightmost point, the area below the lines is called coverage. Now, it's your job to calculate the maximum coverage area, if yi (i = 1, 2 ... n) can be swapped arbitrarily.

    Input

    The first line of the input is an integer t (t <= 100), indicate the number of cases.

    Each case starts with one integer n (2 <= n <= 1000) in a line. Then follows n lines, each consists of two integers x y (1 <= xy <= 105) representing a point.

    Cases are separated by one blank line.

    Output

    For each case, output the answer in a line, keep 1 digit after the decimal point.

    Sample Input

    2
    3
    1 1
    2 2
    3 3
    
    3
    1 2
    4 1
    2 5
    

    Sample Output

    4.5
    10.0
    题意:对于题目给的点,x固定,而与x组合的y可以任意交换,求如何安置y可使这些点组成线段下面的面积最大,最大面积是多少



    #include <iostream>
    #include <algorithm>
    #include <stdio.h>
    using namespace std;
    typedef long long ll;
    
    const int mod = 1e9 + 7;
    const int maxn = 1010;
    
    int main()
    {
    	int t;
    	scanf("%d", &t);
    	while (t--)
    	{
    		int n;
    		double x[maxn], y[maxn], dis[maxn];
    		scanf("%d", &n);
    		for (int i = 0; i < n; i++)
    		{
    			scanf("%lf%lf", x + i, y + i);
    		}
    		sort(x, x + n);
    		sort(y, y + n);
    		dis[0] = x[1] - x[0];
    		dis[n - 1] = x[n - 1] - x[n - 2];
    		for (int i = 1; i < n - 1; i++)
    		{
    			dis[i] = x[i + 1] - x[i - 1];
    		}
    		sort(dis, dis + n);
    		double ans = 0;
    		for (int i = 0; i < n; i++)
    		{
    			ans += dis[i] * y[i];
    		}
    		printf("%.1f
    ", ans / 2.0);
    	}
    	return 0;
    }




    Fighting~
  • 相关阅读:
    汉语-词语:慧根
    汉语-词语:悟性
    汉语-汉字:耑
    汉语-汉字:羣
    汉语-词语:觉察
    汉语-词语:醒悟
    汉语-词语:觉悟
    Redhat 5禁止IPv6
    从缓存行出发理解volatile变量、伪共享False sharing、disruptor
    【故障处理141119】一次数据库不繁忙时一条sql语句2个运行计划导致业务超时的故障处理
  • 原文地址:https://www.cnblogs.com/Archger/p/8451588.html
Copyright © 2011-2022 走看看