ACdream 1135(MST-最小生成树边上2个值，维护第一个最小的前提下让另一个最小)

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

Sample Output

3 3

直接把边按l从小到大（第一关键字），c从小到大（第二关键字）排序,然后用Kruskal算法

#include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<functional> #include<iostream> #include<cmath> #include<cctype> #include<ctime> using namespace std; #define For(i,n) for(int i=1;i<=n;i++) #define Fork(i,k,n) for(int i=k;i<=n;i++) #define Rep(i,n) for(int i=0;i<n;i++) #define ForD(i,n) for(int i=n;i;i--) #define RepD(i,n) for(int i=n;i>=0;i--) #define Forp(x) for(int p=pre[x];p;p=next[p]) #define Lson (x<<1) #define Rson ((x<<1)+1) #define MEM(a) memset(a,0,sizeof(a)); #define MEMI(a) memset(a,127,sizeof(a)); #define MEMi(a) memset(a,128,sizeof(a)); #define INF (2139062143) #define F (1000000007) #define MAXN (1000+10) long long mul(long long a,long long b){return (a*b)%F;} long long add(long long a,long long b){return (a+b)%F;} long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;} typedef long long ll; int n; char a[MAXN][MAXN]; ll p10[MAXN]={0}; ll pow2(ll b) { if (b==1) return 10; if (b==0) return 1; if (p10[b]) return p10[b]; ll p=pow2(b/2)%F; p=(p*p)%F; if (b&1) { p=(p*10)%F; } p10[b]=p; return p; } ll pow2(ll a,ll b) { if (b==1) return a; if (b==0) return 1; ll p=pow2(a,b/2)%F; p=p*p%F; if (b&1) { p=(p*a)%F; } return p; } ll tot[MAXN]={0}; ll mulinv(ll a) { return pow2(a,F-2); } int main() { // freopen("sum.in","r",stdin); // freopen("sum.out","w",stdout); scanf("%d",&n); For(i,n) { scanf("%s",a[i]+1); } /* For(i,n) { For(j,n) cout<<a[i][j]; cout<<endl; } */ For(i,n) { For(j,n) tot[i]+=a[i][j]-'0'+a[j][i]-'0'; } // For(i,n) cout<<tot[i]<<endl; // cout<<mul(pow2(10,1232),mulinv(pow2(10,1232)))<<endl; // cout<<mulinv(9); ll c9=mulinv(9); For(i,n) p10[i]=pow2(i); ll ans=0; For(i,n) { ll t=sub(p10[n-i+1],1),a=tot[i]; t=mul(t,c9); t=mul(a,t); ans=add(ans,mul(t,i)); } cout<<ans<<endl; return 0; }