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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 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.
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 |
Output for 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 |
Case 1: 4 0 3 7 7 Case 2: 4 0 3 Impossible Impossible |
Dataset is huge, user faster I/O methods.
/* *********************************************** Author :guanjun Created Time :2016/6/6 18:42:59 File Name :1002.cpp ************************************************ */ #include <iostream> #include <cstring> #include <cstdlib> #include <stdio.h> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <iomanip> #include <list> #include <deque> #include <stack> #define ull unsigned long long #define ll long long #define mod 90001 #define INF 0x3f3f3f3f #define maxn 10010 #define cle(a) memset(a,0,sizeof(a)) const ull inf = 1LL << 61; const double eps=1e-5; using namespace std; priority_queue<int,vector<int>,greater<int> >pq; struct Node{ int x,y; }; struct cmp{ bool operator()(Node a,Node b){ if(a.x==b.x) return a.y> b.y; return a.x>b.x; } }; bool cmp(int a,int b){ return a>b; } int edge[510][510]; int w[510][510]; int vis[510],lowcost[510]; int n,m,k; int dis[maxn]; void prime(){ memset(dis,-1,sizeof dis); dis[k]=0; cle(vis); vis[k]=1; for(int i=0;i<n;i++){ lowcost[i]=edge[k][i]; } int Min,x; while(1){ Min=INF; for(int i=0;i<n;i++){ if(lowcost[i]<Min&&!vis[i]){ Min=lowcost[i],x=i; } } if(Min==INF)break; vis[x]=1; dis[x]=Min; for(int i=0;i<n;i++){ if(!vis[i]&&max(dis[x],edge[x][i])<lowcost[i]){ lowcost[i]=max(edge[x][i],dis[x]); } } } } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif //freopen("out.txt","w",stdout); int T,x,y,z; cin>>T; for(int t=1;t<=T;t++){ printf("Case %d:\n",t); memset(edge,INF,sizeof edge); cin>>n>>m; for(int i=1;i<=m;i++){ scanf("%d%d%d",&x,&y,&z); if(z<edge[x][y]){ edge[x][y]=z; edge[y][x]=z; } } cin>>k; prime(); for(int i=0;i<n;i++){ if(dis[i]==-1)puts("Impossible"); else printf("%d\n",dis[i]); } } return 0; }
Lightoj 1002 - Country Roads（prim算法）
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原文地址：http://www.cnblogs.com/pk28/p/5564871.html
