UVA 1001 Say Cheese(奶酪里的老鼠)（flod）

题意：无限大的奶酪里有n(0<=n<=100)个球形的洞。你的任务是帮助小老鼠A用最短的时间到达小老鼠O所在位置。奶酪里的移动速度为10秒一个单位，但是在洞里可以瞬间移动。洞和洞可以相交。输入n个球的位置和半径，以及A和O的坐标，求最短时间。

分析：

1、因为洞可以相交，所以在计算两点距离时要判断一下if(dist > num[i].r + num[j].r)。

2、两球间的距离为球心间距离-两球半径，起点和终点不是球，可将半径设为0。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 100 + 10;
const int MAXT = 10000 + 10;
using namespace std;
double d[MAXN][MAXN];
struct Point{
    int x, y, z, r;
    void read(){
        scanf("%d%d%d", &x, &y, &z);
    }
}num[MAXN];
double getD(Point &A, Point &B){
    return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y) + (A.z - B.z) * (A.z - B.z));
}
int main(){
    int n;
    int kase = 0;
    while(scanf("%d", &n) == 1){
        if(n == -1) return 0;
        memset(d, 0, sizeof d);
        for(int i = 0; i < n; ++i){
            num[i].read();
            scanf("%d", &num[i].r);
        }
        num[n].read(), num[n].r = 0;
        num[n + 1].read(), num[n + 1].r = 0;
        for(int i = 0; i < n + 2; ++i){
            for(int j = i + 1; j < n + 2; ++j){
                double dist = getD(num[i], num[j]);
                if(dist > num[i].r + num[j].r){
                    d[i][j] = d[j][i] = dist - num[i].r - num[j].r;
                }
                else{
                    d[i][j] = d[j][i] = 0;
                }
            }
        }
        for(int k = 0; k < n + 2; ++k){
            for(int i = 0; i < n + 2; ++i){
                for(int j = 0; j < n + 2; ++j){
                    d[i][j] = Min(d[i][j], d[i][k] + d[k][j]);
                }
            }
        }
        printf("Cheese %d: Travel time = %.0lf sec\n", ++kase, d[n][n + 1] * 10);
    }
    return 0;
}