CSU - 1548 三分找最值

最新推荐文章于 2021-06-24 09:52:14 发布

June.lu

最新推荐文章于 2021-06-24 09:52:14 发布

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本文链接：https://blog.youkuaiyun.com/Turne/article/details/50209977

解决从坐标(0,0)到(x,y)的最短路径问题，需要考虑河流的存在并决定是修建道路还是桥梁以达到最低成本。通过将河流合并、使用三分查找优化搜索过程，实现最优解。

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Design road

Time Limit: 2000MS

Memory Limit: 262144KB

64bit IO Format: %lld & %llu

Submit Status

Description

You need to design road from (0, 0) to (x, y) in plane with the lowest cost. Unfortunately, there are N Rivers between (0, 0) and (x, y).It costs c1 Yuan RMB per meter to build road, and it costs c2 Yuan RMB per meter to build a bridge. All rivers are parallel to the Y axis with infinite length.

Input

There are several test cases.
Each test case contains 5 positive integers N,x,y,c1,c2 in the first line.(N ≤ 1000,1 ≤ x,y≤ 100,000,1 ≤ c1,c2 ≤ 1000).
The following N lines, each line contains 2 positive integer xi, wi ( 1 ≤ i ≤ N ,1 ≤ xi ≤x, xi-1+wi-1 < xi , xN+wN ≤ x),indicate the i-th river(left bank) locate xi with wi width.
The input will finish with the end of file.

Output

For each the case, your program will output the least cost P on separate line, the P will be to two decimal places .

Sample Input

1 300 400 100 100
100 50
1 150 90 250 520
30 120

Sample Output

50000.00
80100.00

Hint

Source

题意：有河流的地方需要建桥每米花费c2，没有的需要建路每米花费c1,使得从（0,0）到（x,y)花费最少。

思路：合并所有河流记为sum，然后在[0,y]里面通过三分寻找一个值y1，使得从（0,0）到（sum，y1）加上（sum，y1）到（x，y）的花费最少。（把河流和道路分开易于寻找最值）

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
using namespace std;
#define eps 1e-3
double distance(double x1,double y1,double x2,double y2)//求两点之间的距离
{
	return sqrt((double)(x1-x2)*(x1-x2) + (double)(y1-y2)*(y1-y2));
}
int main()
{
#ifdef CDZSC_June
	freopen("t.txt","r",stdin);
#endif
	int n;
	double x,y,c1,c2,s,w,sum;
	while(~scanf("%d%lf%lf%lf%lf",&n,&x,&y,&c1,&c2))
	{
		sum = 0;
		for(int i =0; i<n; i++)
		{
			scanf("%lf%lf",&s,&w);
			sum += w;//合并河流
		}
		sum = x - sum;//获取需要建路的宽度
		double L = 0,R = y,mid,midmid;
		while(R - L > eps)
		{
			mid = (L+R)/2;
			midmid = (mid+R)/2;
			double s1 = distance(0,0,sum,mid)*c1+distance(sum,mid,x,y)*c2;
			double s2 =  distance(0,0,sum,midmid)*c1+distance(sum,midmid,x,y)*c2;
			//midmid必定大于mid，所以当s1>s2时，意味着最小值在靠近s2的那一边，所以令L = mid来缩小最小值的寻找范围，反之同理
			if(s1 > s2){
					L = mid;
			}
			else{
				R = midmid;
			}
		}
		printf("%.2f\n",distance(0,0,sum,mid)*c1+distance(sum,mid,x,y)*c2);
	}
	return 0;
}