HDU1245:Saving James Bond(Floyd)

Problem Description

This time let us consider the situation in the movie "Live and Let Die" in which James Bond, the world's most famous spy, was captured by a group of drug dealers. He was sent to a small piece of land at the center of a lake filled with crocodiles. There he performed the most daring action to escape -- he jumped onto the head of the nearest crocodile! Before the animal realized what was happening, James jumped again onto the next big head... Finally he reached the bank before the last crocodile could bite him (actually the stunt man was caught by the big mouth and barely escaped with his extra thick boot).
Assume that the lake is a 100×100 square one. Assume that the center of the lake is at (0,0) and the northeast corner at (50,50). The central island is a disk centered at (0,0) with the diameter of 15. A number of crocodiles are in the lake at various positions. Given the coordinates of each crocodile and the distance that James could jump, you must tell him whether he could escape.If he could,tell him the shortest length he has to jump and the min-steps he has to jump for shortest length.

 

Input

The input consists of several test cases. Each case starts with a line containing n <= 100, the number of crocodiles, and d > 0, the distance that James could jump. Then one line follows for each crocodile, containing the (x, y) location of the crocodile. Note that x and y are both integers, and no two crocodiles are staying at the same position.  

 

Output

For each test case, if James can escape, output in one line the shortest length he has to jump and the min-steps he has to jump for shortest length. If it is impossible for James to escape that way, simply ouput "can't be saved".

 

Sample Input

   
   
   
   

    
    
    
    4 10
17 0
27 0
37 0
45 0
1 10
20 30
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    42.50 5
can't be saved
   
   
   
   

 

 题意：小道以原点为圆心，直径15米，然后给出所有鳄鱼的坐标，问007能否逃出生天

思路：我们只需要求出每两个鳄鱼间的距离，就可以转化为求最短路的情况了，由于精度问题，只能用C++过，GC++过不了

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
const double inf = 100000000;
double map[105][105];
int s[105],e[105],step[105][105],len1,len2;

struct node
{
    double x,y;
} a[105];

void floyd(int n)
{
    int i,j,k;
    for(k = 0; k<=n; k++)
        for(i = 0; i<=n; i++)
            for(j = 0; j<=n; j++)
                if(map[i][j]>map[i][k]+map[k][j])
                {
                    map[i][j]=map[i][k]+map[k][j];
                    step[i][j]=step[i][k]+step[k][j];
                }
}

int main()
{
    int n,i,j,k,len;
    double d,x,y;
    while(~scanf("%d%lf",&n,&d))
    {
        len = 1;
        for(i = 1; i<=n; i++)
        {
            scanf("%lf%lf",&x,&y);
            if(fabs(x)<=7.5 && fabs(y)<=7.5)//只将所有在小岛外的点存入图中
                continue;
            a[len].x = x;
            a[len++].y = y;
        }
        n = len;
        if(n == 1)//只存了一个点，直接判断
        {
            if(d>=42.5)
                printf("42.50 1\n");
            else
                printf("can't be saved\n");
            continue;
        }
        for(i = 0; i<=n; i++)
            for(j = 0; j<=n; j++)
            {
                map[i][j] = inf;
                step[i][j] = 0;
            }
        for(i = 1; i<n; i++)
        {
            for(j = 1; j<n; j++)
            {
                if(j==i)
                {
                    map[i][j] = 0;
                    continue;
                }
                map[i][j] = sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y));//枚举所有距离
                step[i][j] = 1;
                if(map[i][j]>d)//步伐不能到达，将该店变为无穷大
                {
                    map[i][j] = inf;
                    step[i][j] = 0;
                }
            }
        }
        len1 = len2 = 0;
        for(i = 1; i<n; i++)
        {
            if(sqrt(a[i].x*a[i].x+a[i].y*a[i].y)<=7.5+d)//起始点必须是从小岛上出发，步伐能到达的点
                s[len1++] = i;
            if((50+a[i].x)<=d || (50-a[i].x)<=d || (50+a[i].y)<=d || (50-a[i].y)<=d)//借速点市所有能到达岸边的点
                e[len2++] = i;
        }
        for(i = 0; i<len1; i++)
        {
            map[0][s[i]] = map[s[i]][0] = sqrt(a[s[i]].x*a[s[i]].x+a[s[i]].y*a[s[i]].y)-7.5;//小岛到起始点的距离
            step[0][s[i]] = step[s[i]][0] = 1;
        }
        for(i = 0; i<len2; i++)
        {
            map[e[i]][n] = map[n][e[i]] = min(min(50+a[e[i]].x,50-a[e[i]].x),min(50+a[e[i]].y,50-a[e[i]].y));//结束点到岸边的最短距离
            step[e[i]][n] = step[n][e[i]] = 1;
        }
        floyd(n);
        if(map[0][n] == inf)
            printf("can't be saved\n");
        else
            printf("%.2f %d\n",map[0][n],step[0][n]);
    }

    return 0;
}