uva 1001 奶酪里的老鼠（狄杰斯特拉或弗洛伊德）

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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

本来不难的一道题，因为图论做得少所以不太熟悉浪费了很多时间，今晚怒改一遍，试了弗洛伊德和狄杰斯特拉算法，都ac了，主要是读题不认真，泪目，今天下午选拔赛两道题也是因为各种数据类型的细节浪费时间，以后要多注意，after all ，detail is devil

#include<cstdio>  #include<cstring>  #include<cmath>  #include<cstdlib>  #include<iostream>  #include<algorithm>  #include<vector>  #include<map>  #include<queue>  #include<stack> #include<string>#include<map> #include<set>using namespace std;  #define LL long long  double INF = 100000000000.0; const int maxn = 100 + 5;struct Node{int x, y, z, r;}nodes[maxn];int vis[maxn];double G[maxn][maxn], d[maxn];int n;void dijkstra(int s) {for(int i = 0; i <= n + 1; i++) d[i] = (i == s ? 0.0 : INF);memset(vis, 0, sizeof(vis));for(int i = 0; i <= n + 1; i++) {double mind = INF;int tag;for(int j = 0; j <= n + 1; j++) {if(!vis[j] && d[j] < mind) {mind = d[j];tag = j;} }vis[tag] = 1;for(int j = 0; j <= n + 1; j++) d[j] = min(d[j], d[tag] + G[tag][j]);//printf("%d %lf %lf\n",tag,mind,d[tag]);}} int main() {//freopen("input.txt", "r", stdin);int hole,kase=0;while(scanf("%d", &hole) && hole != -1){n = hole;int x, y, z, r;for(int i = 1; i <= hole; i++){scanf("%d%d%d%d", &x, &y, &z, &r);nodes[i].x = x;nodes[i].y = y;nodes[i].z = z;nodes[i].r = r;}scanf("%d%d%d", &x, &y, &z);nodes[0].x = x;nodes[0].y = y;nodes[0].z = z;nodes[0].r = 0;scanf("%d%d%d", &x, &y, &z);nodes[n + 1].x = x;nodes[n + 1].y = y;nodes[n + 1].z = z;nodes[n + 1].r = 0;for(int i = 0; i <= n + 1; i++) G[i][i] = 0;for(int i = 0; i < n + 1; i++)for(int j = i + 1; j <= n +1; j++) if(i != j){ Node& u = nodes[i]; Node& v = nodes[j];G[i][j] = G[j][i] = max(0.0, sqrt( (double)pow(u.x - v.x, 2) + (double)pow(u.y - v.y, 2) + (double)pow(u.z - v.z, 2) ) - (double)u.r - (double)v.r);}//dijkstra(0);//printf("%lf\n", G[0][1]);/* for(int k=0;k<=n + 1;k++)          for(int i=0;i<=n + 1;i++)           for(int j=0;j<=n + 1;j++)          G[i][j]=min(G[i][j],G[i][k]+G[k][j]);*///int ans = (int)(G[0][n+1] * 10+ 0.5);dijkstra(0);int ans = (int)(d[n + 1] * 10 + 0.5);//printf("Cheese %d: Travel time = %d sec\n", ++kase, ans);printf("Cheese %d: Travel time = %d sec\n", ++kase, ans);}return 0;}

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