HDU 1245 Saving James Bond(floyd)

Saving James Bond

#include <cstdio>
#include <cstring>
#include <cmath>
#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++) {//建立起各鳄鱼之间的路径
            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)//借助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;
}