题目描述
Since childhood you have been fascinated by model railroads. Designing your own tracks,complicated intersections, train stations with miniatures of travellers, train operators, luggage is so much fun! However, it also needs a lot of space. Since your house is more than full by now, you decide to move to the garden.
You have already moved all your completed tracks outside when you notice an important flaw:
Since different tracks were in different rooms before, there are stations which cannot be reached from each other. That has to change!
Since you have already fixed the exact positions of the stations, you know the lengths for all possible connections you can build and also which stations are connected already. All connections can be used in both directions. You can decide to remove some existing connections and instead build new ones of at most the same total length. Is it possible to rearrange the railroads so that every station is reachable from all other stations?
输入
The input consists of:
• one line with three integers n (2 ≤ n ≤ 5 · 104 ), m (0 ≤ m ≤ 2.5 · 105 ) and l (0 ≤ l ≤ m),where n is the number of stations, m is the number of possible connections and l is the number of connections already built;
• m lines describing the connections. Each connection is described by:
– one line with three integers a, b (1 ≤ a, b ≤ n), and c (0 ≤ c ≤ 5 · 103 ) describing that there is a connection from station a to station b of length c.
The first l of those connections already exist.
输出
Output “possible” if it is possible to construct a connected network as described above,otherwise output “impossible”.
样例输入
4 6 3
1 2 1
2 1 2
3 4 2
1 3 2
1 4 3
2 4 2
样例输出
possible
提示
Figure E.1 depicts the first sample case. It is possible to connect all stations by removing the connections between stations 1 and 2 of length 2 and instead building the connection between stations 2 and 4. The curvature of the rails does not matter because you have a hammer. In the second case, depicted in Figure E.2, it is not possible to connect all three stations.
题意： 有n个节点，m条边，其中前L条边是已经存在的，已经存在的L条边能被等长度的一些边替换，即删除一些已存在边，可以新建出m条边中一些等长的边。问是否能用已有长度联通所有的N个节点。
题解： 因为给出了前L条边是已经存在的，首先就对其求和。所以拆掉一些较长边，新建一些较短边，只要使其联通就好了，而不是用较短边通过组合代替较长边。
可能他给的前I条就是不联通的，但是前I条如果够长，还是能新建删除来联通的，如果一开始给的I条边的sum就特别小，那联通的边长一定会超sum，就是impossible.
然后因为是所有点都联通，那么直接求一颗最小生成树。然后对比生成树长度是否小于或等于前L条边总长度即可得到答案。
#include<bits/stdc++.h>
#define LL long long
#define M(a,b) memset(a,b,sizeof a)
#define pb(x) push_back(x)
using namespace std;
const int maxn=5e4+7;
struct edge
{
int from,to;
LL val;
edge(){}
edge(int a,int b,LL c)
{
from=a,to=b,val=c;
}
bool operator <(const edge &a)const
{
return val<a.val;
}
}mp[maxn*10];
int n,m,l,z[maxn];
int finds(int x)
{
return z[x]==x?x:z[x]=finds(z[x]);
}
void join(int x,int y)
{
z[finds(x)]=y;
}
int main()
{
scanf("%d%d%d",&n,&m,&l);
for(int i=1;i<=n;i++)z[i]=i;
LL sum=0;
for(int i=1;i<=m;i++)
{
scanf("%d%d%lld",&mp[i].from,&mp[i].to,&mp[i].val);
sum+=i<=l?mp[i].val:0;
}
sort(mp+1,mp+1+m);
int cnt=0;
LL dist=0;
for(int i=1;i<=m;i++)
{
if(finds(mp[i].from)!=finds(mp[i].to))
{
join(mp[i].from,mp[i].to);
dist+=mp[i].val;
cnt++;
}
}
printf("%s
",cnt==n-1&&dist<=sum?"possible":"impossible");
}