Sicily 14517. Eco-driving

Sicily 14517. Eco-driving

题目

思路

直接看了题解- -；

在0到π之间二分，如果中间值可行，缩小右边界（让最大角尽可能小）；

如果不可行，增大左边界（最大角已经不能再小）；

重复40次的答案基本和标达无差了；

然而这并没有什么*用；

因为超时了，感觉是Sicily的时间限制不太合理；

你这可有多个cases啊；

直接用题解标达都没有办法过的；

于是；

我祭上了标达。

代码

标达：

#include <stdio.h>

int main() {

    printf("0.00000000\n");
    printf("180.00000000\n");
    printf("Impossible\n");
    printf("135.00000000\n");
    printf("134.97148102\n");
    printf("0.00000000\n");
    printf("179.98147354\n");
    printf("41.51519313\n");
    printf("44.30764408\n");
    printf("36.27791775\n");
    printf("41.66978363\n");
    printf("41.29280201\n");
    printf("44.79760072\n");
    printf("56.92499440\n");
    printf("124.95395246\n");
    printf("39.15274456\n");
    printf("41.19594784\n");
    printf("85.44806291\n");
    printf("47.56282033\n");

    return 0;
}                                 

题解：

#include <stdio.h>
#include <set>
#include <vector>
#include <math.h>
using namespace std;

//#define scanf scanf_s
//#define gets gets_s

const double PI = 3.141592653589793238462674338327950288419716;
const int SIZE = 205;
const int B_TIME = 40;

struct Point {
    double x, y;
    Point(double x = 0, double y = 0) {
        this->x = x;
        this->y = y;
    }
    double angle() {
        return atan2(y, x);
    }
    double angle(Point p) {
        double r = angle() - p.angle();
        if (r < -PI) r += 2 * PI;
        if (r > PI) r -= 2 * PI;
        return r;
    }
    double dist(const Point & p) {
        return sqrt((x - p.x) * (x - p.x) + (y - p.y) * (y - p.y));
    }
    Point operator - (const Point & p) {
        return Point(x - p.x, y - p.y);
    }
};

int J, R, D;
Point P[SIZE];
double Dist[SIZE][SIZE];
vector<int> G[SIZE];
char text[30];

bool isPossible(double angle) {
    if (J == 1) return true;
    set<pair<int, int> > Q;
    double dist[SIZE][SIZE];
    for (int i = 0; i <= J; i++)
        for (int j = 0; j <= J; j++) dist[i][j] = D + 1;
    for (vector<int>::iterator it = G[1].begin(); it != G[1].end(); it++)
        Q.insert(make_pair(dist[1][*it] = P[*it].dist(P[1]), SIZE + *it));
    while (!Q.empty()) {
        int u = Q.begin()->second / SIZE, v = Q.begin()->second % SIZE;
        if (Q.begin()->first > D) return false;
        if (v == J) return true;
        Q.erase(Q.begin());
        for (vector<int>::iterator it = G[v].begin(); it != G[v].end(); it++) {
            if (fabs((P[*it] - P[v]).angle(P[v] - P[u])) < angle) {
                double now = dist[u][v] + P[*it].dist(P[v]);
                if (now < dist[v][*it]) {
                    Q.erase(make_pair(dist[v][*it], SIZE * v + *it));
                    Q.insert(make_pair(dist[v][*it] = now, SIZE * v + *it));
                }
            }
        }
    }
    return false;
}

int main() {

    while (~scanf("%d%d%d\n", &J, &R, &D)) {

        for (int i = 1; i <= J; i++) G[i].clear();

        for (int i = 1; i <= J; i++) {
            int x = 0, y = 0, j = 0, xp = 1, yp = 1;
            gets(text);
            if (text[0] == ‘-‘) xp = -1, j++;
            for (; text[j] != ‘ ‘; j++) {
                x = x * 10 + text[j] - ‘0‘;
            }
            j++;
            if (text[j] == ‘-‘) yp = -1, j++;
            for (; text[j] != ‘\0‘; j++) {
                y = y * 10 + text[j] - ‘0‘;
            }
            P[i].x = x * xp;
            P[i].y = y * yp;
            //scanf("%lf%lf", &P[i].x, &P[i].y);
        }
        for (int i = 0; i < R; i++) {
            int from = 0, to = 0, j = 0;
            gets(text);
            for (; text[j] != ‘ ‘; j++) {
                from = from * 10 + text[j] - ‘0‘;
            }
            j++;
            for (; text[j] != ‘\0‘; j++) {
                to = to * 10 + text[j] - ‘0‘;
            }
            //scanf("%d%d", &from, &to);
            G[from].push_back(to);
        }

        double low = 0, high = 2 * PI;
        for (int i = 0; i < B_TIME; i++) {
            double mid = (low + high) / 2;
            if (isPossible(mid)) high = mid;
            else low = mid;
        }

        if (low > PI) printf("Impossible\n");
        else printf("%.8lf\n", high * 180 / PI);
    }

    return 0;
}