UVa 1001 Say Cheese (Dijkstra)

题意：给定一个三维空间的一些球和起始位置和结束位置，问你最短要花的时间是多少。

析：建图，所有的位置都建立图，边权就是距离，最小求一次最短路即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<double, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-5;
const int maxn = 100 + 10;
const int mod = 1e6;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
struct Point{
  double x, y, z, r;
};
Point a[maxn];
double G[maxn][maxn];
double d[maxn];
double distant(int i, int j){
  return max(0.0, 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) + (a[i].z-a[j].z)*(a[i].z-a[j].z)) - a[i].r - a[j].r);
}

double dijkstra(int s, int t){
  priority_queue<P, vector<P>, greater<P> >pq;
  fill(d, d+n+5, INF);
  d[s] = 0.0;
  pq.push(P(0.0, s));

  while(!pq.empty()){
    P p = pq.top();  pq.pop();
    if(p.second == t)  return p.first;
    int u = p.second;
    if(d[u] < p.first)  continue;
    for(int i = 0; i < n+2; ++i){
      if(d[i] > d[u] + G[u][i]){
        d[i] = d[u] + G[u][i];
        pq.push(P(d[i], i));
      }
    }
  }
}

int main(){
  int kase = 0;
  while(scanf("%d", &n) == 1 && n >= 0){
    for(int i = 0; i < n; ++i)
      scanf("%lf %lf %lf %lf", &a[i].x, &a[i].y, &a[i].z, &a[i].r);
    int s = n, t = n+1;
    for(int i = n; i < n+2; ++i)  scanf("%lf %lf %lf", &a[i].x, &a[i].y, &a[i].z);
    a[n].r = a[n+1].r = 0.0;
    for(int i = 0; i < n+2; ++i)
      for(int j = i+1; j < n+2; ++j)
        G[i][j] = G[j][i] = distant(i, j);
    double ans = dijkstra(s, t) * 10;
    printf("Cheese %d: Travel time = %.f sec
", ++kase, ans);
  }
  return 0;
}