Abandoned country
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1756 Accepted Submission(s):
475
Problem Description
An abandoned country has n(n≤100000)
villages which are numbered from 1 to n.
Since abandoned for a long time, the roads need to be re-built. There are m(m≤1000000)
roads to be re-built, the length of each road is wi(wi≤1000000).
Guaranteed that any two wi
are different. The roads made all the villages connected directly or indirectly
before destroyed. Every road will cost the same value of its length to rebuild.
The king wants to use the minimum cost to make all the villages connected with
each other directly or indirectly. After the roads are re-built, the king asks a
men as messenger. The king will select any two different points as starting
point or the destination with the same probability. Now the king asks you to
tell him the minimum cost and the minimum expectations length the messenger will
walk.
Input
The first line contains an integer T(T≤10)
which indicates the number of test cases.
For each test case, the first line contains two integers n,m indicate the number of villages and the number of roads to be re-built. Next m lines, each line have three number i,j,wi, the length of a road connecting the village i and the village j is wi.
For each test case, the first line contains two integers n,m indicate the number of villages and the number of roads to be re-built. Next m lines, each line have three number i,j,wi, the length of a road connecting the village i and the village j is wi.
Output
output the minimum cost and minimum Expectations with
two decimal places. They separated by a space.
Sample Input
1
4 6
1 2 1
2 3 2
3 4 3
4 1 4
1 3 5
2 4 6
Sample Output
6 3.33
Author
HIT
Source
Recommend
这题我完全没有思路，比赛时也不会写，看了别人的代码，貌似懂了一点点。。
题意：给了一个图，求最小生成树和期望值，最小生成树指所有的点连接成一条完整的路的最小权值和，期望指图中任意两个点的距离之和除以所有的边的个数（如测试数据：4个点，就是用1-2,1-3,1-4,2-3,2-4,3-4这六条距离的和除以边长总数6）。
附上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #define M 200010 6 #define ll long long 7 using namespace std; 8 ll n,m; 9 10 struct edge 11 { 12 int u,v; 13 int w; 14 } e[M*5]; 15 16 struct node 17 { 18 int u,v,w; 19 int ans; ///ans记录经过这条边能达到的边的个数 20 int next; ///next记录上一次连接的边的序号 21 }ee[M]; 22 23 int father[M],head[M]; ///father数组表示根节点，head记录每条边最后一次连接的边的序号 24 ll ans,sum; 25 int tol; 26 27 ll dfs(int u,int p) ///搜索 28 { 29 ll num=0; 30 for(int i=head[u];i!=-1;i=ee[i].next) 31 { 32 if(ee[i].v!=p) 33 { 34 ee[i].ans+=dfs(ee[i].v,ee[i].u); 35 sum+=ee[i].ans*(n-ee[i].ans)*ee[i].w; ///此边能到达的边的个数*能到此边的边的个数*边的长度 36 // cout<<ee[i].ans<<" "<<n-ee[i].ans<<" "<<ee[i].w<<endl; 37 //cout<<sum<<endl; 38 num+=ee[i].ans; 39 } 40 } 41 return num; 42 } 43 44 void add_node(edge a) ///建立邻接表 45 { 46 ee[tol].u=a.u; ///正向 47 ee[tol].v=a.v; 48 ee[tol].w=a.w; 49 ee[tol].ans=1; 50 ee[tol].next=head[a.u]; 51 head[a.u]=tol++; 52 ee[tol].u=a.v; ///反向 53 ee[tol].v=a.u; 54 ee[tol].w=a.w; 55 ee[tol].ans=1; 56 ee[tol].next=head[a.v]; 57 head[a.v]=tol++; 58 } 59 60 int find(int x) ///搜索根节点 61 { 62 while(x!=father[x]) 63 x=father[x]; 64 return x; 65 } 66 67 void sourch(int x,int y,int z,edge ss) 68 { 69 x=find(x); 70 y=find(y); 71 if(x!=y) ///若根节点不相同 72 { 73 add_node(ss); ///并且此边存在，建立邻接表 74 father[x]=y; ///将其连接在一起 75 ans+=z; 76 } 77 78 } 79 80 void init() 81 { 82 ans=0,sum=0; 83 int i,j; 84 for(i=1; i<=n; i++) 85 { 86 father[i]=i; ///将所有父节点初始定义为自己本身 87 head[i]=-1; 88 } 89 tol=0; 90 for(i=0; i<m; i++) 91 { 92 sourch(e[i].u,e[i].v,e[i].w,e[i]); ///求最小生成树 93 } 94 dfs(1,-1); 95 ll nn=n*(n-1)/2; 96 printf("%I64d %.2lf ",ans,(double)sum/nn); 97 } 98 99 bool cmp(edge a,edge b) 100 { 101 return a.w<b.w; 102 } 103 104 int main() 105 { 106 int T,i; 107 scanf("%d",&T); 108 while(T--) 109 { 110 scanf("%lld%lld",&n,&m); 111 for(i=0; i<m; i++) 112 { 113 scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w); ///记录每边的数据 114 } 115 sort(e,e+m,cmp); ///要选取最小的花费，因此要进行排序 116 init(); 117 } 118 return 0; 119 }