dfs用来求这个节点属于哪一块 用floyd求出各点间的最短距离 再求出每个节点可达到的最远距离MAX[i]以及这两个块中的最长距离a,b. 枚举当两个节点不在一块中 dis = max{(dis[i][j]+MAX[i]+MAX[j] ),a,b),在求出dis中最小的。
View Code 1 /* 2 ID: shangca2 3 LANG: C++ 4 TASK: cowtour 5 */ 6 #include <iostream> 7 #include<cstdio> 8 #include<cstring> 9 #include<algorithm> 10 #include<stdlib.h> 11 #include<cmath> 12 using namespace std; 13 #define INF 0x3f3f3f 14 struct node 15 { 16 int x,y; 17 }no[200]; 18 double w[200][200],m[200],a[5]; 19 int n,vis[200]; 20 void dfs(int x,int d) 21 { 22 int i; 23 for(i = 1; i <= n ; i++) 24 if(!vis[i]&&w[x][i]!=INF) 25 { 26 vis[i] = d; 27 dfs(i,d); 28 } 29 } 30 int main() 31 { 32 freopen("cowtour.in","r",stdin); 33 freopen("cowtour.out","w",stdout); 34 int i,j,k; 35 char c; 36 cin>>n; 37 for(i = 1; i <= n ; i++) 38 cin>>no[i].x>>no[i].y; 39 for(i = 1; i <= n ; i++) 40 { 41 for(j = 1; j <= n ; j++) 42 { 43 cin>>c; 44 if(c=='1') 45 w[i][j] = 1.0*sqrt((no[i].x-no[j].x)*(no[i].x-no[j].x)+(no[i].y-no[j].y)*(no[i].y-no[j].y)); 46 else 47 w[i][j] = INF; 48 } 49 } 50 int g = 1; 51 for(i = 1; i <= n ; i++) 52 if(!vis[i]) 53 { 54 vis[i] = g; 55 dfs(i,g++); 56 } 57 for(i = 1; i <= n ; i++) 58 w[i][i] = 0; 59 for(i = 1; i <= n ; i++) 60 for(j = 1; j <= n ; j++) 61 for(k = 1; k <= n ; k++) 62 { 63 if(w[j][k]>w[j][i]+w[i][k]) 64 w[j][k]=w[j][i]+w[i][k]; 65 } 66 for(i = 1; i <= n ; i++) 67 for(j = 1; j <= n ;j++) 68 { 69 if(w[i][j]<INF) 70 a[vis[i]] = max(a[vis[i]],w[i][j]); 71 } 72 for(i = 1; i <= n ; i++) 73 { 74 m[i] = 0; 75 for(j = 1; j <= n ; j++) 76 if(w[i][j]<INF) 77 m[i] = max(m[i],w[i][j]); 78 } 79 double mi=INF,mm = INF; 80 81 for(i = 1; i <= n ; i++) 82 { 83 for(j = 1; j <= n ; j++) 84 { 85 if(vis[i]!=vis[j]) 86 { 87 mi = m[i]+m[j]+1.0*sqrt((no[i].x-no[j].x)*(no[i].x-no[j].x)+(no[i].y-no[j].y)*(no[i].y-no[j].y)); 88 mi = max(mi,a[1]); 89 mi = max(mi,a[2]); 90 } 91 if(mi<mm) 92 mm = mi; 93 } 94 } 95 printf("%.6lf\n",mm); 96 return 0; 97 }