uva 1001（最短路）

题意：在一个三维的奶酪里面有n（n<=100）个洞，老鼠A想到达老鼠B的位置，在洞里面可以瞬间移动，在洞外面的移动速度为10秒一个单位，求最短时间

题解：如果两个洞相交，那么d[i][j]=0；如果不相交，那么d[i][j]=dist-(r[i]+r[j]),dist为这两个洞圆心之间的欧几里得距离

再用Dijkstra处理就可以了

#include <cstdio>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
#define ll long long
#define _cle(m, a) memset(m, a, sizeof(m))
#define repu(i, a, b) for(int i = a; i < b; i++)
#define repd(i, a, b) for(int i = b; i >= a; i--)
#define sfi(n) scanf("%d", &n)
#define sfd(n) scanf("%lf", &n)
#define pfi(n) printf("%d\n", n)
#define pfd(n) printf("%lf\n", n)
#define sfi2(n, m) scanf("%d%d", &n, &m)
#define sfd2(n, m) scanf("%lf%lf", &n, &m)
#define pfi2(n, m) printf("%d %d\n", n, m)
#define pfi3(a, b, c) printf("%d %d %d\n", a, b, c)
#define maxn 105
double dd[maxn][maxn];
struct Cir{
   double x, y, z, r;
}c[maxn];
double get_d(Cir c1, Cir c2)
{
    return sqrt((c1.x - c2.x) * (c1.x - c2.x) +
                (c1.y - c2.y) * (c1.y - c2.y) +
                (c1.z - c2.z) * (c1.z - c2.z));
}

const double inf = 1000000000000.0;
struct Dijkstra {
    int n;    //图,须手动传入！
    double E[maxn][maxn], d[maxn];
    int p[maxn];    //最短路径，父亲

    void init(int n) {
        this->n = n;
        repu(i, 0, n) repu(j, 0, n)
        E[i][j] = dd[i][j];
    }

    void solve(int s) {
        static bool vis[maxn];

        memset(vis, 0, sizeof(vis));
        memset(p, 255, sizeof(p));
        repu(i, 0, n + 1) d[i] = inf;
        d[s] = 0.0;
        //repu(i, 0, n) pfd(d[i]);
        while(1) {
            int u = -1;
            for(int i = 0; i < n; i ++) {
                if(!vis[i] && (u == -1||d[i] < d[u])) {
                    u = i;
                }
            }
            if(u == -1 || d[u] == inf) break;
            vis[u] = true;
            for(int v = 0; v < n; v ++) {
                if(d[u] + E[u][v] < d[v]) {
                    d[v] = d[u] + E[u][v];
                    p[v] = u;
                }
            }
        }
    }
} dij;

int main()
{
    int n;
    int kase = 0;
    while(~sfi(n) && n != -1)
    {
        kase++;
        repu(i, 1, n + 1)
        sfd2(c[i].x, c[i].y), sfd2(c[i].z, c[i].r);

        sfd2(c[0].x, c[0].y), sfd(c[0].z);
        sfd2(c[n + 1].x, c[n + 1].y), sfd(c[n + 1].z);
        c[0].r = c[n + 1].r = 0.0;

        repu(i, 0, n + 2)
        repu(j, 0, n + 2)
        {
           dd[i][j] = get_d(c[i], c[j]) - (c[i].r + c[j].r);
           if(dd[i][j] < 0.0) dd[i][j] = 0.0;
           //printf("%d %d %lf\n", i, j, dd[i][j]);
        }

        dij.init(n + 2);
        dij.solve(0);

        //repu(i, 0, n + 2) pfd(dij.d[i]);
        printf("Cheese %d: Travel time = %d sec\n",
                     kase, (int)round(dij.d[n + 1] * 10.0));
    }
    return 0;
}