题目：https://vjudge.net/problem/POJ-1125
Stockbrokers are known to overreact to rumours. You have been contracted to develop a method of spreading disinformation amongst the stockbrokers to give your employer the tactical edge in the stock market. For maximum effect, you have to spread the rumours in the fastest possible way.
Unfortunately for you, stockbrokers only trust information coming from their "Trusted sources" This means you have to take into account the structure of their contacts when starting a rumour. It takes a certain amount of time for a specific stockbroker to pass the rumour on to each of his colleagues. Your task will be to write a program that tells you which stockbroker to choose as your starting point for the rumour, as well as the time it will take for the rumour to spread throughout the stockbroker community. This duration is measured as the time needed for the last person to receive the information.Input
Each person is numbered 1 through to the number of stockbrokers. The time taken to pass the message on will be between 1 and 10 minutes (inclusive), and the number of contacts will range between 0 and one less than the number of stockbrokers. The number of stockbrokers will range from 1 to 100. The input is terminated by a set of stockbrokers containing 0 (zero) people.
Output
It is possible that your program will receive a network of connections that excludes some persons, i.e. some people may be unreachable. If your program detects such a broken network, simply output the message "disjoint". Note that the time taken to pass the message from person A to person B is not necessarily the same as the time taken to pass it from B to A, if such transmission is possible at all.
Sample Input
3 2 2 4 3 5 2 1 2 3 6 2 1 2 2 2 5 3 4 4 2 8 5 3 1 5 8 4 1 6 4 10 2 7 5 2 0 2 2 5 1 5 0
Sample Output
3 2
题意：
这题每个人传递消息是同时进行的，求最后一个人知道消息的最短时间。
先枚举每个人为起点的情况，然后这个情况下求每个人知道消息的最短时间，然后遍历一遍找到最大的那个，就是最后一个人收到消息的时间了。然后比较出所以情况中最后一个人知道消息的时间的最小值，即所求答案。
思路：
枚举每个人为起点的最短路，求最小值，迪杰斯特拉算法直接暴力过，，
代码：
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<map> #define maxn 10005 #define inf 1e9+7 using namespace std; int dis[maxn]; int mp[maxn][maxn]; int n,mazz; void init() { for(int i=1;i<=n;i++) { dis[i]=inf; } } void dijkstra(int n,int s) { bool vis[maxn]; memset(vis,false,sizeof(vis)); for(int i=1;i<=n;i++) { dis[i]=mp[s][i]; } dis[s]=0; vis[s]=true; for(int i=1;i<=n;i++){ int u=s; mazz=inf; for(int j=1;j<=n;j++) { if(!vis[j]&&dis[j]<mazz) { u=j; mazz=dis[j]; } } vis[u]=true; for(int zz=1;zz<=n;zz++) { if(dis[zz]>dis[u]+mp[u][zz]&&!vis[zz]) { dis[zz]=dis[u]+mp[u][zz]; } } } } int main() { while(scanf("%d",&n)) { if(n==0)break; int i,j,m; for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { if(i<=j) mp[i][j]=mp[j][i]=inf; } } for(i=1;i<=n;i++) { scanf("%d",&m); for(j=1;j<=m;j++) { int x,y; scanf("%d%d",&x,&y); mp[i][x]=y; } } init(); int mazz=inf,k=0; for(i=1;i<=n;i++) { dijkstra(n,i); int ma=0; for(j=1;j<=n;j++) { if(j!=i) { if(dis[j]>ma){ma=dis[j];} } } if(ma<mazz){mazz=ma;k=i;} } if(k!=0) printf("%d %d\n",k,mazz); else printf("disjoint\n"); } return 0; }
水题一道，，套套板子
