題目：

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, outp**ut 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

題意:鱷魚的坐標，n是鱷魚的數(shù)目，d是一步跳的最遠距離
題目輸出他走到終點需要的最小距離，還有最小的距離中走的最小的步數(shù)

參考代碼：
https://blog.csdn.net/libin56842/article/details/17034011
實現(xiàn)起來有點崩潰

#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)//只將所有在小島外的點存入圖中continue;a[len].x = x;a[len++].y = y;}n = len;if(n == 1)//只存了一個點，直接判斷{if(d>=42.5)printf("42.50 1\n");elseprintf("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));//枚舉所有距離step[i][j] = 1;if(map[i][j]>d)//步伐不能到達，將該店變?yōu)闊o窮大{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)//起始點必須是從小島上出發(fā)，步伐能到達的點s[len1++] = i;if((50+a[i].x)<=d || (50-a[i].x)<=d || (50+a[i].y)<=d || (50-a[i].y)<=d)//借速點市所有能到達岸邊的點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;//小島到起始點的距離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));//結(jié)束點到岸邊的最短距離step[e[i]][n] = step[n][e[i]] = 1;}floyd(n);if(map[0][n] == inf)printf("can't be saved\n");elseprintf("%.2f %d\n",map[0][n],step[0][n]);}return 0; }

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