CodeChef

题目链接

 

AMRExchange is the latest cryptocurrency exchange that has become very popular among cryptocurrency traders.

On AMRExchange, there are N cryptocurrencies - let us denote the i^th currency by C_i. Mpairs of these cryptocurrencies are tradable - one unit of currency C_x can be converted to one unit of currency C_y with risk C_xy.

Mr. X, an avid cryptocurrency collector, wants to start with 1 unit of any of the Ncryptocurrencies and perform a sequence of trades. He wants to do it in such a way that for each of the N cryptocurrencies, there was at least one point during the trading sequence during which he held one unit of that cryptocurrency.

The overall risk of the sequence of trades is the maximum risk in the sequence of trades. Minimize the overall risk with which Mr. X can achieve this. Print "Impossible" if no such sequence of trades is possible.

Input

• The first line contains a single integer T - the total number of testcases.
• Each testcase is of the following format:
  • First line contains 2 space-separated integers - N and M. N denotes the number of cryptocurrencies, M denotes the number of tradable ordered cryptocurrency pairs.
  • M lines follow. Each line contains 3 space-separated positive integers - C_x, C_yand C_xy. This line denotes that one unit of currency C_x can be converted into one unit of currency C_y with risk C_xy.

Output

• For each testcase, print the minimum overall risk with which Mr. X can own at least one unit of each cryptocurrency at some point in time.
• If it is not possible for Mr. X to achieve this, then print “Impossible”.

Constraints

• 1 ≤ T ≤ 5
• 1 ≤ N, M ≤ 10⁵
• 1 ≤ C_x, C_y ≤ N
• 1 ≤ C_xy ≤ 10⁹

Example

Input
2
3 6
1 2 1
2 3 3
3 1 3
1 3 1
3 2 1
3 2 5
4 3
1 2 1
2 3 1
2 4 1

Output
1
Impossible

Explanation

Testcase 1: Mr. X can start with cryptocurrency C₁ and follow the following sequence to minimize overall risk:

• Convert C₁ to C₃ with risk 1.
• Convert C₃ to C₂ with risk 1.

 

The overall risk would be 1.

Testcase 2: There are a total of 6 sequences of trades are possible, and none of them satisfy our property. We list them here:

Starting with C₁:

• C₁ -> C₂ -> C₃
• C₁ -> C₂ -> C₄

In the first sequence, Mr. X won't be able to own C₄ because units of C₃ cannot be converted to units of C₄. Similarly, in the second sequence, Mr. X won't be able to own C₃ because units of C₄ cannot be converted to units of C₃.

Starting with C₂:

• C₂ -> C₃
• C₂ -> C₄

Starting with C₃:

• C₃ (cannot be converted to any other cryptocurrency)

Starting with C₄:

• C₄ (cannot be converted to any other cryptocurrency)

Hence, there is no possible sequence using which Mr. X can own one unit of all cryptocurrencies at some point in time.

题意 

给一个有边权的图，找出最小的权值，使得以这个权值能走遍所有的点。

分析 

二分权值。对于某个权值，以这个权值为基准来缩点，构建出新的简单有向图，然后求出拓扑序，检查是否可以形成链状。

 

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include <queue>
#include <vector>
#include<bitset>
#include<map>
#include<deque>
#include<stack>
using namespace std;
typedef pair<int,int> pii;
#define X first
#define Y second
#define pb push_back
#define mp make_pair
#define ms(a,b) memset(a,b,sizeof(a))
const int inf = 0x3f3f3f3f;
const int maxn = 1e5+5;
const int mod = 77200211+233;
#define lson l,m,2*rt
#define rson m+1,r,2*rt+1
typedef long long ll;
vector<pii> G[maxn];
vector<int> g[maxn];
int indeg[maxn];
int n,m;
int pre[maxn],lowlink[maxn],sccno[maxn],dfs_block,scc_cnt;
stack<int> S;
void dfs(int u,int lim){
    pre[u]=lowlink[u]=++dfs_block;
    S.push(u);
    for(int i=0;i<G[u].size();i++){
        if(G[u][i].Y>lim) continue;
        int v=G[u][i].X;
        if(!pre[v]){
            dfs(v,lim);
            lowlink[u]=min(lowlink[u],lowlink[v]);
        }else if(!sccno[v]){
            lowlink[u]=min(lowlink[u],pre[v]);
        }
    }
    if(lowlink[u]==pre[u]){
        scc_cnt++;
        while(1){
            int x=S.top();S.pop();
            sccno[x]=scc_cnt;
            if(x==u) break;
        }
    }
    return;
}
void find_scc(int n,int lim){
    dfs_block=scc_cnt=0;
    ms(sccno,0);
    ms(pre,0);
    for(int i=1;i<=n;i++){
        if(!pre[i]) dfs(i,lim);
    }
    return;
}

bool check(int mid){
    ms(indeg,0);
    for(int i=0;i<=scc_cnt;i++){
        g[i].clear();
    }
    find_scc(n,mid);
    for(int i=1;i<=n;i++){
        for(int j=0;j<G[i].size();j++){
            int u,v,w;
            u=i;
            v=G[i][j].X;
            w=G[i][j].Y;
            if(w>mid) continue;
            if(sccno[u]==sccno[v]) continue;
            g[sccno[u]].pb(sccno[v]);
            indeg[sccno[v]]++;
        }
    }

    queue<int> que;
    for(int i=1;i<=scc_cnt;i++) if(indeg[i]==0) que.push(i);
    vector<int> topu;
    while(!que.empty()){
        int now=que.front();
        que.pop();
        topu.pb(now);
        for(int i=0;i<g[now].size();i++){
            indeg[g[now][i]]--;
            if(indeg[g[now][i]]==0){
                que.push(g[now][i]);
            }
        }
    }

    if(topu.size()<=1) return true;
    for(int i=0;i<topu.size()-1;i++){
        int now=topu[i];
        bool flag=0;
        for(int j=0;j<g[now].size();j++){
            int tmp=g[now][j];
            if(tmp==topu[i+1]){
                flag=1;
                break;
            }
        }
        if(!flag){
            return false;
        }
    }
    return true;
}
void solve(){
    int l=1,r=1e9+5;
    int mid;
    int ans=inf;
    while(l<=r){
        mid=(l+r)>>1;
        if(check(mid)){
            r=mid-1;
            ans=min(mid,ans);
        }else l=mid+1;
    }
    if(ans==inf) puts("Impossible");
    else printf("%d
",ans);
}

int main(){
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif // LOCAL
    int t,u,v,w;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        for(int i=0;i<=n;i++) G[i].clear();
        for(int i=0;i<m;i++){
            scanf("%d%d%d",&u,&v,&w);
            G[u].pb(mp(v,w));
        }
        solve();
    }
    return 0;
}