UVa1001 Say Cheese (最短路,Dijkstra)

链接：http://vjudge.net/problem/UVA-1001

分析：将起点A和终点O也看作洞，把半径r设为0就可以了，然后预处理出n+2个结点间的距离，从起点A开始用Dijkstra跑一边最短路后，最短距离乘以移动速度即可。

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <queue>
 5 using namespace std;
 6
 7 typedef long long LL;
 8 const int maxn = 100 + 5;
 9 const int INF = 1e29;
10
11 int n;
12 int x[maxn], y[maxn], z[maxn], r[maxn], vis[maxn];
13 double dist[maxn][maxn], d[maxn];
14
15 struct HeapNode {
16     double d;
17     int u;
18     bool operator < (const HeapNode& rhs) const {
19         return d > rhs.d;
20     }
21 };
22
23 LL dijkstra(int s) {
24     memset(vis, 0, sizeof(vis));
25     for (int i = 0; i <= n + 1; i++) d[i] = INF;
26     priority_queue<HeapNode> q;
27     d[0] = 0.0;
28     q.push((HeapNode){0.0, 0});
29     while (!q.empty()) {
30         int u = q.top().u; q.pop();
31         if (vis[u]) continue; vis[u] = 1;
32         for (int i = 0; i <= n + 1; i++)
33             if (d[i] > d[u] + dist[u][i]) {
34                 d[i] = d[u] + dist[u][i];
35                 q.push((HeapNode){d[i], i});
36             }
37     }
38     return round(d[n + 1] * 10) + 0.5;
39 }
40
41 int main() {
42     int kase = 0;
43     while (scanf("%d", &n) == 1 && n != -1) {
44         for (int i = 1; i <= n; i++) scanf("%d%d%d%d", &x[i], &y[i], &z[i], &r[i]);
45         scanf("%d%d%d", &x[0], &y[0], &z[0]); r[0] = 0;
46         scanf("%d%d%d", &x[n + 1], &y[n + 1], &z[n + 1]); r[n + 1] = 0;
47         for (int i = 0; i <= n + 1; i++)
48             for (int j = 0; j <= n + 1; j++) {
49                 if (i == j) { dist[i][j] = 0.0; continue; }
50                 dist[i][j] = sqrt(pow(x[i] - x[j], 2) + pow(y[i] - y[j], 2) + pow(z[i] - z[j], 2));
51                 if (dist[i][j] <= r[i] + r[j]) dist[i][j] = 0.0;
52                 else dist[i][j] = dist[i][j] - (r[i] + r[j]);
53             }
54         printf("Cheese %d: Travel time = %lld sec\n", ++kase, dijkstra(0));
55     }
56     return 0;
57 }