Acdream 1135 MST

MstTime limit:2000/1000ms (java/others) Memory limit:128000/64000kb (java/others) Problem Descriptiongiven a connected, undirected graph, a spanning tree of that graph was a subgraph that is a tree and con  Nects all the vertices together. A Single graph can has many different spanning trees. We can also assign a weight to each edge, which was a number representing how unfavorable it was, and use this to assign a W Eight 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.
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Now we do the problem more complex. We assign each edge, 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 have minimal sum of cost. Input

There is multiple test cases.
The first line contains, 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 is an edge with l length and c cos T between A and B.

1 <= N <= 10,000
1 <= M <= 100,000
1 <= A, b <= N
1 <= L, c <= 10,000

Outputfor each test case output, 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 51 2 1 1 2 3 1 13 4 1 11 3 1 22 4 1 3

Sample Output

3 3

sourcedut200901102managerdut200901102 Problem Solving: Yes, yes, it's MST, just a double-keyword sort.

1#include <bits/stdc++.h>2 using namespacestd;3typedefLong LongLL;4 Const intMAXN =500010;5 structarc{6     intU,v,length,cost;7     BOOL operator< (ConstArc &rhs)Const{8         if(length = = rhs.length)returnCost <Rhs.cost;9         returnLength <rhs.length;Ten     } One }E[MAXN]; A intUF[MAXN]; - intFind (intx) { -     if(x = uf[x]) uf[x] =Find (uf[x]); the     returnUf[x]; - } - intMain () { -     intn,m; +      while(~SCANF ("%d%d",&n,&m)) { -          for(inti =0; I < m; ++i) +scanf"%d%d%d%d",&e[i].u,&e[i].v,&e[i].length,&e[i].cost); A          for(inti =0; I <= N; ++i) Uf[i] =i; atSort (e,e +m); -LL length =0, cost =0, cnt =0; -          for(inti =0; I < m && CNT +1< n; ++i) { -             intU =Find (e[i].u); -             intv =Find (E[I].V); -             if(U = = v)Continue; inUf[u] =v; -Length + =e[i].length; toCost + =E[i].cost; +++CNT; -         } the         if(CNT +1= = N) printf ("%lld%lld\n", length,cost); *         ElsePuts"-1-1"); $     }Panax Notoginseng     return 0; -}

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Acdream 1135 MST