zoukankan      html  css  js  c++  java
  • YTU 2547: Repairing a Road

    2547: Repairing a Road

    时间限制: 1 Sec  内存限制: 128 MB
    提交: 3  解决: 2

    题目描述

    You live in a small town with R bidirectional roads connecting C crossings and you want to go from crossing 1 to crossing C as soon as possible. You can visit other crossings before arriving at crossing C, but it’s not mandatory.

    You have exactly one chance to ask your friend to repair exactly one existing road, from the time you leave crossing 1. If he repairs the i-th road for t units of time, the crossing time after that would be viai-t. It's not difficult to see that it takes vi units of time to cross that road if your friend doesn’t repair it.

    You cannot start to cross the road when your friend is repairing it.

    Input

    There will be at most 25 test cases. Each test case begins with two integers C and R (2<=C<=100, 1<=R<=500). Each of the next R lines contains two integers xiyi (1<=xiyi<=C) and two positive floating-point numbers vi and ai (1<=vi<=20,1<=ai<=5), indicating that there is a bidirectional road connecting crossing xi and yi, with parameters vi and ai (see above). Each pair of crossings can be connected by at most one road. The input is terminated by a test case with C=R=0, you should not process it.

    Output

    For each test case, print the smallest time it takes to reach crossing C from crossing 1, rounded to 3 digits after decimal point. It’s always possible to reach crossing C from crossing 1.

    输入

    输出

    样例输入

    3 21 2 1.5 1.82 3 2.0 1.52 11 2 2.0 1.80 0

    样例输出

    2.5891.976

    迷失在幽谷中的鸟儿,独自飞翔在这偌大的天地间,却不知自己该飞往何方……

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <iostream>
    #include <cmath>
    using namespace std;
    const int MAXN = 110;
    const int MAXE = 1010;
    const double EPS = 1e-6;
    inline int sgn(double x)
    {
        if(fabs(x) < EPS) return 0;
        return x > 0 ? 1 : -1;
    }
    double fpai(double t, double v, double a)
    {
        return 1 - log(a) * v * pow(a, - t);
    }
    inline void update_min(double &a, const double &b)
    {
        if(a > b) a = b;
    }
    double mat[MAXN][MAXN];
    int x[MAXE], y[MAXE];
    double v[MAXE], a[MAXE];
    int n, m;
    void floyd()
    {
        for(int k = 1; k <= n; ++k)
            for(int i = 1; i <= n; ++i)
                for(int j = 1; j <= n; ++j) update_min(mat[i][j], mat[i][k] + mat[k][j]);
    }
    double find_t(int i, int x, int y, double l, double r)
    {
        double L = l, R = r;
        while(R - L > EPS)
        {
            double mid = (L + R) / 2;
            if(fpai(mid, v[i], a[i]) > 0) R = mid;
            else L = mid;
        }
        if(sgn(fpai(L, v[i], a[i])) != 0) return l;
        return L;
    }
    double solve()
    {
        double t, ans = mat[1][n];
        for(int i = 0; i < m; ++i)
        {
            t = find_t(i, x[i], y[i], mat[1][x[i]], ans);
            update_min(ans, t + v[i] * pow(a[i], -t) + mat[y[i]][n]);
            t = find_t(i, y[i], x[i], mat[1][y[i]], ans);
            update_min(ans, t + v[i] * pow(a[i], -t) + mat[x[i]][n]);
        }
        return ans;
    }
    int main()
    {
        while(scanf("%d%d", &n, &m) != EOF)
        {
            if(n == 0 && m == 0) break;
            for(int i = 1; i <= n; ++i)
            {
                for(int j = 1; j <= n; ++j) mat[i][j] = 1e5;
                mat[i][i] = 0;
            }
            for(int i = 0; i < m; ++i)
            {
                int aa, bb;
                double cc;
                scanf("%d%d%lf%lf", &aa, &bb, &cc, &a[i]);
                x[i] = aa;
                y[i] = bb;
                v[i] = cc;
                update_min(mat[aa][bb], cc);
                update_min(mat[bb][aa], cc);
            }
            floyd();
            printf("%.3f
    ", solve());
        }
    }

  • 相关阅读:
    26、面向对象设计模式之单例模式——泛型单例
    Unity 汽车碰撞
    makeObjectsPerformSelector对数组中的对象发送消息执行对象中方法
    NSHashTable NSPointerArray
    webrtc 音频一点相关知识
    记一次ios加急上架经历
    iOS 获取当前正在显示的ViewController
    ios表单上传图片或文件
    https适配
    swift block
  • 原文地址:https://www.cnblogs.com/im0qianqian/p/5989484.html
Copyright © 2011-2022 走看看