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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.
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.
For each the case, your program will output the least cost P on separate line, the P will be to two decimal places .
1 300 400 100 100
100 50
1 150 90 250 520
30 120
50000.00 80100.00
题意:有一个人要从(0,0)->(x,y)，中间有n条河,每条河都有一个宽度和一个起始位置,并且上一条河与下一条河无交点,修一公里路花费 c1 ，一公里河花费 c2，问人从起点到终点的最小费用?
题解:我们将所有的和平移到最右边,那么这里就只要枚举河岸就OK了,单峰极值,三丰求解。(0,0)->(X,t)为河,(X,t)->(x,y)为路.花费为 sqrt(X*X+t*t)*c2+sqrt((x-X)*(x-X)+(y-t)*(y-t))*c1
#include<stdio.h> #include<iostream> #include<string.h> #include<math.h> #include<algorithm> using namespace std; int n; double x,y,c1,c2,sum,X; const double eps = 1e-8; double Calc(double t) { return sqrt(X*X+t*t)*c2+sqrt((x-X)*(x-X)+(y-t)*(y-t))*c1; } double solve(double MIN,double MAX) { double Left, Right; double mid, midmid; double mid_value, midmid_value; Left = MIN; Right = MAX; while (Left +eps < Right) { mid = (Left + Right) / 2; midmid = (mid + Right) / 2; mid_value = Calc(mid); midmid_value = Calc(midmid); if (mid_value <= midmid_value) Right = midmid; else Left = mid; } return Left; } int main() { while(scanf("%d%lf%lf%lf%lf",&n,&x,&y,&c1,&c2)!=EOF) { X = 0; for(int i=1; i<=n; i++) { double a,b; scanf("%lf%lf",&a,&b); X+=b; } double k = solve(0,y); printf("%.2lf\n",Calc(k)); } }
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原文地址：http://www.cnblogs.com/liyinggang/p/5792248.html
