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;

}