HDU 1245 Saving James Bond

大意：在一片海域海域（长、宽为100的正方形）中有一孤岛，孤岛是直径为15.0的圆形区域，在孤岛的周围有一些点，问你是否能够成功的跳出这片海域，最少的时间是多少以及步数是多少？

思路：关键在于建图，图中的点分3类：1、孤岛到点。2、点到点。3、点到终点。

注意，点到终点的距离是50分别减去X、Y坐标的最小值，孤岛到点的距离需要减去半径。还有如果可以一步跳出去，就一步跳出去。建完图后，bfs或者Dijkstra都行。

 

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
#include <algorithm>
#include <map>
#include <set>
using namespace std;

const int maxn = 110;
const double eps = 1e-6;
const double INF = 1e60;

struct node
{
    double x, y;
}A[maxn];

int dcmp(double x) { if(fabs(x) < eps) return 0; else return x < 0? -1 : 1; }

double sqr(double x) { return x*x; }

double Dist1(node a) { return fabs(sqrt(sqr(a.x) + sqr(a.y)) - 7.50) ; } //到点的距离
double Dist2(node a, node b) { return sqrt(sqr(a.x-b.x) + sqr(a.y-b.y)) ; } //同类点之间的距离 
double Dist3(node a) { return dcmp(fabs(50-fabs(a.x)) - fabs(50-fabs(a.y))) > 0 ? fabs(50-fabs(a.y)) : fabs(50-fabs(a.x)); }
 //到终点的距离
  
int n, D;

double d[maxn][maxn];
double cost[maxn];
int step[maxn];

void read_case()
{
    for(int i = 1; i <= n; i++) scanf("%lf%lf", &A[i].x, &A[i].y);
}

void build()
{
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= n; j++) d[i][j] = Dist2(A[i], A[j]);
    
    for(int i = 1; i <= n; i++)
    {
        d[0][i] = Dist1(A[i]);
        d[i][0] = d[n+1][i] = INF;
        d[i][n+1] = Dist3(A[i]);
    }
    d[0][n+1] = INF;
}

void bfs(int s, int t)
{
    int cur, next;
    queue<int> q;
    for(int i = 0; i <= n+1; i++) cost[i] = (i == s)? 0:INF;
    step[0] = 0;
    q.push(s);
    while(!q.empty())
    {
        cur = q.front(); q.pop();
        for(next = 1; next <= n+1; next++)
        {
            if(dcmp(d[cur][next] - D) <= 0 && 
              (dcmp(cost[cur] + d[cur][next] - cost[next]) < 0 || 
              (dcmp(cost[cur] + d[cur][next] - cost[next]) <= 0 && dcmp(step[cur]+1 < step[next]) < 0)))
              { //能使得距离减少或者在距离相等的情况下使得步数减少则入队 
                    cost[next] = cost[cur] + d[cur][next];
                    step[next] = step[cur]+1;
                    q.push(next);
              }
        }
    }
}

void solve()
{
    read_case();
    if(dcmp(D-42.50) >= 0) { printf("42.5 1\n"); return ; } //一步跳出去 
    build();
    bfs(0, n+1);
    if(dcmp(cost[n+1] - INF) < 0) printf("%.2lf %d\n", cost[n+1], step[n+1]);
    else printf("can't be saved\n");
}

int main()
{
    while(~scanf("%d%d", &n, &D))
    {
        solve();
    }
    return 0;
}