UVa1001 Say Cheese (最短路,Dijkstra)
Posted
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了UVa1001 Say Cheese (最短路,Dijkstra)相关的知识,希望对你有一定的参考价值。
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 }
以上是关于UVa1001 Say Cheese (最短路,Dijkstra)的主要内容,如果未能解决你的问题,请参考以下文章
UVA1001 Say Cheese(Dijkstra或Floyd)
UVA - 1001 Say Cheese(奶酪里的老鼠)(flod)
UVa 1001 奶酪里的老鼠(Dijkstra或Floyd)
UVa1048 Low Cost Air Travel--最短路