2017多校训练Contest5: 1006 Rikka with Graph hdu6090
2017-08-20 17:51
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Problem Description
As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:
For an undirected graph G with n nodes
and m edges,
we can define the distance between (i,j) (dist(i,j))
as the length of the shortest path between i and j.
The length of a path is equal to the number of the edges on it. Specially, if there are no path between i and j,
we make dist(i,j) equal
to n.
Then, we can define the weight of the graph G (wG)
as ∑ni=1∑nj=1dist(i,j).
Now, Yuta has n nodes,
and he wants to choose no more than m pairs
of nodes (i,j)(i≠j) and
then link edges between each pair. In this way, he can get an undirected graph G with n nodes
and no more than m edges.
Yuta wants to know the minimal value of wG.
It is too difficult for Rikka. Can you help her?
In the sample, Yuta can choose (1,2),(1,4),(2,4),(2,3),(3,4).
Input
The first line contains a number t(1≤t≤10),
the number of the testcases.
For each testcase, the first line contains two numbers n,m(1≤n≤106,1≤m≤1012).
Output
For each testcase, print a single line with a single number -- the answer.
Sample Input
1
4 5
Sample Output
14
考虑直接以一个点为中心练成一个放射状的图
多出来的边再把发散出来的点两两连线
维护下每种情况的贡献即可
#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cassert>
#include<iostream>
#include<algorithm>
using namespace std;
int a[100001];
int main()
{
int T;
scanf("%d",&T);
while(T>0)
{
T--;
long long n,m;
scanf("%I64d%I64d",&n,&m);
if(m>=n*(n-1LL)/2LL)
{
printf("%I64d\n",n*(n-1LL));
continue;
}
long long ans=(n-1LL)*n*n;
long long i;
if(m<=n-1LL)
{
long long d=0;
for(i=1;i<=m;i++)
{
ans-=2LL*((n-1LL)+d*(n-2LL));
d++;
}
}
else
{
long long d=0;
for(i=1;i<=n-1LL;i++)
{
ans-=2LL*((n-1LL)+d*(n-2LL));
d++;
}
m-=(n-1LL);
ans-=2LL*m;
}
printf("%I64d\n",ans);
}
return 0;
}
As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:
For an undirected graph G with n nodes
and m edges,
we can define the distance between (i,j) (dist(i,j))
as the length of the shortest path between i and j.
The length of a path is equal to the number of the edges on it. Specially, if there are no path between i and j,
we make dist(i,j) equal
to n.
Then, we can define the weight of the graph G (wG)
as ∑ni=1∑nj=1dist(i,j).
Now, Yuta has n nodes,
and he wants to choose no more than m pairs
of nodes (i,j)(i≠j) and
then link edges between each pair. In this way, he can get an undirected graph G with n nodes
and no more than m edges.
Yuta wants to know the minimal value of wG.
It is too difficult for Rikka. Can you help her?
In the sample, Yuta can choose (1,2),(1,4),(2,4),(2,3),(3,4).
Input
The first line contains a number t(1≤t≤10),
the number of the testcases.
For each testcase, the first line contains two numbers n,m(1≤n≤106,1≤m≤1012).
Output
For each testcase, print a single line with a single number -- the answer.
Sample Input
1
4 5
Sample Output
14
考虑直接以一个点为中心练成一个放射状的图
多出来的边再把发散出来的点两两连线
维护下每种情况的贡献即可
#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cassert>
#include<iostream>
#include<algorithm>
using namespace std;
int a[100001];
int main()
{
int T;
scanf("%d",&T);
while(T>0)
{
T--;
long long n,m;
scanf("%I64d%I64d",&n,&m);
if(m>=n*(n-1LL)/2LL)
{
printf("%I64d\n",n*(n-1LL));
continue;
}
long long ans=(n-1LL)*n*n;
long long i;
if(m<=n-1LL)
{
long long d=0;
for(i=1;i<=m;i++)
{
ans-=2LL*((n-1LL)+d*(n-2LL));
d++;
}
}
else
{
long long d=0;
for(i=1;i<=n-1LL;i++)
{
ans-=2LL*((n-1LL)+d*(n-2LL));
d++;
}
m-=(n-1LL);
ans-=2LL*m;
}
printf("%I64d\n",ans);
}
return 0;
}
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