I am going to my home. There are many cities and many bi-directional roads between them. The cities are numbered from 0 to n-1 and each road has a cost. There are m roads. You are given the number of my city t where I belong. Now from each city you have to find the minimum cost to go to my city. The cost is defined by the cost of the maximum road you have used to go to my city.
For example, in the above picture, if we want to go from 0 to 4, then we can choose
1) 0 - 1 - 4 which costs 8, as 8 (1 - 4) is the maximum road we used
2) 0 - 2 - 4 which costs 9, as 9 (0 - 2) is the maximum road we used
3) 0 - 3 - 4 which costs 7, as 7 (3 - 4) is the maximum road we used
So, our result is 7, as we can use 0 - 3 - 4.
Input
Input starts with an integer T (≤ 20), denoting the number of test cases.
Each case starts with a blank line and two integers n (1 ≤ n ≤ 500) and m (0 ≤ m ≤ 16000). The next m lines, each will contain three integers u, v, w (0 ≤ u, v < n, u ≠ v, 1 ≤ w ≤ 20000) indicating that there is a road between u and v with cost w. Then there will be a single integer t (0 ≤ t < n). There can be multiple roads between two cities.
Output
For each case, print the case number first. Then for all the cities (from 0 to n-1) you have to print the cost. If there is no such path, print ‘Impossible’.
Sample Input
2
5 6
0 1 5
0 1 4
2 1 3
3 0 7
3 4 6
3 1 8
1
5 4
0 1 5
0 1 4
2 1 3
3 4 7
1
Sample Output
Case 1:
4
0
3
7
7
Case 2:
4
0
3
Impossible
Impossible
最短路SPFA另用;
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#pragma comment(linker,"/STACK:102400000")
#define WW freopen("output.txt","w",stdout)
const int Max = 160000;
struct node
{
int v;
int w;
int next;
}Edge[Max*10];
int Head[550];
int n,m;
int top;
int t;
int Dis[550];
bool vis[550];
void Build(int u,int v,int w)
{
Edge[top].v=v;
Edge[top].w=w;
Edge[top].next=Head[u];
Head[u]=top;
top++;
}
void SPFA()
{
int a,b;
memset(Dis,INF,sizeof(Dis));
memset(vis,false,sizeof(vis));
Dis[t]=0;
vis[t]=true;
queue<int>Q;
Q.push(t);
while(!Q.empty())
{
a=Q.front();
Q.pop();
b=Head[a];
while(b!=-1)
{
if(Dis[Edge[b].v]>max(Dis[a],Edge[b].w))
{
Dis[Edge[b].v]=max(Dis[a],Edge[b].w);
if(!vis[Edge[b].v])
{
vis[Edge[b].v]=true;
Q.push(Edge[b].v);
}
}
b=Edge[b].next;
}
vis[a]=false;
}
}
int main()
{
int T;
int u,v,w;
int W=1;
scanf("%d",&T);
while(T--)
{
scanf("%d %d",&n,&m);
memset(Head,-1,sizeof(Head));
top=0;
for(int i=0;i<m;i++)
{
scanf("%d %d %d",&u,&v,&w);
Build(u,v,w);
Build(v,u,w);
}
scanf("%d",&t);
SPFA();
printf("Case %d:
",W++);
for(int i=0;i<n;i++)
{
if(Dis[i]<INF)
{
printf("%d
",Dis[i]);
}
else
{
printf("Impossible
");
}
}
}
return 0;
}
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