MST
Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)
Problem Description
Given a connected, undirected graph, a spanning tree of that graph is a subgraph that is a tree and connects all the vertices together. A single graph can have many different spanning trees. We can also assign a weight to each edge, which is a number representing how unfavorable it is, and use this to assign a weight to a spanning tree by computing the sum of the weights of the edges in that spanning tree. A minimum spanning tree (MST) is then a spanning tree with weight less than or equal to the weight of every other spanning tree.
------ From wikipedia
Now we make the problem more complex. We assign each edge two kinds of
weight: length and cost. We call a spanning tree with sum of length less
than or equal to others MST. And we want to find a MST who has minimal
sum of cost.
Input
There are multiple test cases.
The first line contains two integers N and M indicating the number of vertices and edges in the gragh.
The next M lines, each line contains three integers a, b, l and c indicating there are an edge with l length and c cost between a and b.
1 <= N <= 10,000
1 <= M <= 100,000
1 <= a, b <= N
1 <= l, c <= 10,000
Output
For each test case output two integers indicating the sum of length and cost of corresponding MST.
If you can find the corresponding MST, please output "-1 -1".
Sample Input
4 5 1 2 1 1 2 3 1 1 3 4 1 1 1 3 1 2 2 4 1 3
Sample Output
3 3
Source
dut200901102
Manager
解题：是的，没错，就是MST，只是是双关键字排序
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long LL;
4 const int maxn = 500010;
5 struct arc{
6 int u,v,length,cost;
7 bool operator<(const arc &rhs)const{
8 if(length == rhs.length) return cost < rhs.cost;
9 return length < rhs.length;
10 }
11 }e[maxn];
12 int uf[maxn];
13 int Find(int x){
14 if(x != uf[x]) uf[x] = Find(uf[x]);
15 return uf[x];
16 }
17 int main(){
18 int n,m;
19 while(~scanf("%d%d",&n,&m)){
20 for(int i = 0; i < m; ++i)
21 scanf("%d%d%d%d",&e[i].u,&e[i].v,&e[i].length,&e[i].cost);
22 for(int i = 0; i <= n; ++i) uf[i] = i;
23 sort(e,e + m);
24 LL length = 0,cost = 0,cnt = 0;
25 for(int i = 0; i < m && cnt + 1 < n; ++i){
26 int u = Find(e[i].u);
27 int v = Find(e[i].v);
28 if(u == v) continue;
29 uf[u] = v;
30 length += e[i].length;
31 cost += e[i].cost;
32 ++cnt;
33 }
34 if(cnt + 1 == n) printf("%lld %lld\n",length,cost);
35 else puts("-1 -1");
36 }
37 return 0;
38 }