Dark roads
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1031 Accepted Submission(s): 450
Problem Description
Economic
times these days are tough, even in Byteland. To reduce the operating
costs, the government of Byteland has decided to optimize the road
lighting. Till now every road was illuminated all night long, which
costs 1 Bytelandian Dollar per meter and day. To save money, they
decided to no longer illuminate every road, but to switch off the road
lighting of some streets. To make sure that the inhabitants of Byteland
still feel safe, they want to optimize the lighting in such a way, that
after darkening some streets at night, there will still be at least one
illuminated path from every junction in Byteland to every other
junction.
What is the maximum daily amount of money the government of Byteland can save, without making their inhabitants feel unsafe?
What is the maximum daily amount of money the government of Byteland can save, without making their inhabitants feel unsafe?
Input
The
input file contains several test cases. Each test case starts with two
numbers m and n, the number of junctions in Byteland and the number of
roads in Byteland, respectively. Input is terminated by m=n=0.
Otherwise, 1 ≤ m ≤ 200000 and m-1 ≤ n ≤ 200000. Then follow n integer
triples x, y, z specifying that there will be a bidirectional road
between x and y with length z meters (0 ≤ x, y < m and x ≠ y). The
graph specified by each test case is connected. The total length of all
roads in each test case is less than 231.
Output
For each test case print one line containing the maximum daily amount the government can save.
Sample Input
7 11
0 1 7
0 3 5
1 2 8
1 3 9
1 4 7
2 4 5
3 4 15
3 5 6
4 5 8
4 6 9
5 6 11
0 0
Sample Output
51
Source
所以边的权值-最小生成树的权值
裸题
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<queue> 7 #include<map> 8 #include<set> 9 #include<vector> 10 #include<cstdlib> 11 #include<string> 12 #define eps 0.000000001 13 typedef long long ll; 14 typedef unsigned long long LL; 15 using namespace std; 16 const int N=200000+100; 17 int parent[N]; 18 int ans; 19 int m,n; 20 struct node{ 21 int u,v,w; 22 }a[N]; 23 bool cmp(node x,node y){ 24 return x.w<y.w; 25 } 26 void init(){ 27 for(int i=0;i<=N;i++)parent[i]=i; 28 } 29 int find(int x){ 30 int r=x; 31 while(parent[r]!=r)r=parent[r]; 32 int i=x; 33 int j; 34 while(i!=r){ 35 j=parent[i]; 36 parent[i]=r; 37 i=j; 38 } 39 return r; 40 } 41 void kruskal(){ 42 for(int i=0;i<m;i++){ 43 int x=find(a[i].u); 44 int y=find(a[i].v); 45 if(x!=y){ 46 parent[x]=y; 47 ans=ans+a[i].w; 48 } 49 } 50 } 51 int main(){ 52 while(scanf("%d%d",&n,&m)!=EOF){ 53 init(); 54 if(m==0&&n==0)break; 55 int sum=0; 56 ans=0; 57 for(int i=0;i<m;i++){ 58 scanf("%d%d%d",&a[i].u,&a[i].v,&a[i].w); 59 sum=sum+a[i].w; 60 } 61 //cout<<sum<<endl; 62 sort(a,a+m,cmp); 63 kruskal(); 64 65 //cout<<ans<<endl; 66 printf("%d ",sum-ans); 67 } 68 }