Lightoj1002——Country Roads（最短路变形）

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
Output for Sample Input
1
Case 1:
4
0
3
7
7
Case 2:
4
0
3
Impossible
Impossible

从u到v点有多条路径，路径的花费是权值最长的那段路，而u到v的最短路径就是花费最小的那个。求给出的点到其他点的所有最短路径
设dis保存当前点的最短路，由于题目还是求最短路，所以本质上还是要每次找到个最小的dis，不同的是dis的求法。
另外还要注意有重边

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <cmath>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 25#define Mod 10001using namespace std;int map[1005][1005],vis[1005],dis[1005],n,m;void dijkstra(int s){    int i,j,mi,v;    memset(vis,0,sizeof(vis));    for(i=0; i<n; ++i)    {        dis[i]=map[s][i];    }    dis[s]=0;    vis[s]=1;    for(i=0; i<n; ++i)    {        mi=INF;        for(j=0; j<n; ++j)        {            if(!vis[j]&&dis[j]<mi)            {                mi=dis[j];                v=j;            }        }        vis[v]=1;        for(j=0; j<n; ++j)        {            if(!vis[j]&&map[v][j]<INF)            {                int tmp=max(dis[v],map[v][j]);                dis[j]=min(dis[j],tmp);            }        }    }}int main(){    int t;    scanf("%d",&t);    for(int cas=1; cas<=t; ++cas)    {        scanf("%d%d",&n,&m);        for(int i=0; i<n; ++i)            for(int j=0; j<n; ++j)                map[i][j]=INF;        while(m--)        {            int u,v,w;            scanf("%d%d%d",&u,&v,&w);            if(map[u][v]>w)            {                map[u][v]=w;                map[v][u]=w;            }        }        int s;        scanf("%d",&s);        dijkstra(s);        printf("Case %d:\n",cas);        for(int i=0; i<n; ++i)        {            if(dis[i]<INF)                printf("%d\n",dis[i]);            else                printf("Impossible\n");        }    }    return 0;}