F - 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.
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".
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
图中边有2个值l,c，求关于l的MST,在此基础上求min∑c
直接把边按l从小到大（第一keyword），c从小到大（第二keyword）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;
}