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Language: Default Stockbroker Grapevine
Description 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 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 Input Your program will input data for different sets of stockbrokers. Each set starts with a line with the number of stockbrokers. Following this is a line for each stockbroker which contains the number of people who they have contact with, who these people are, 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 Output For each set of data, your program must output a single line containing the person who results in the fastest message transmission, and how long before the last person will receive any given message after you give it to this person, measured in integer minutes. 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 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 3 10 Source |
题意：在一个有向图中选一个起点，能从这个起点到达所有其他点，还要保证起点到其他点距离中的最大值最小。
思路：先用floyd求出所有两点之间的距离，然后枚举起点，看以这个为起点能不能到达所有点且最大值最小。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 111
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;
int mp[MAXN][MAXN];
int n;
void floyd() //folyd求出所有点之间的最短路
{
int i,j,k;
for (k=1;k<=n;k++)
for (i=1;i<=n;i++)
for (j=1;j<=n;j++)
{
if (i==k||j==k)
continue;
mp[i][j]=min(mp[i][j],mp[i][k]+mp[k][j]);
}
}
int main()
{
int i,j,m,t,e;
while (scanf("%d",&n)&&n)
{
memset(mp,INF,sizeof(mp));
for (i=1;i<=n;i++)
{
scanf("%d",&m);
if (m>0)
{
for (j=1;j<=m;j++)
{
scanf("%d%d",&e,&t);
mp[i][e]=t;
}
}
}
floyd();
int mi=INF;
int p;
int flag=0;
for (i=1;i<=n;i++)//枚举1~n作为起点
{
int ma=-INF;
for (j=1;j<=n;j++)//判断以i作为起点能不能从i走到其他所有点
if (i!=j)
{
if (mp[i][j]==INF)
break;
ma=max(ma,mp[i][j]);//找出以i为起点的情况下最远的点，记录下长度
}
if (j==n+1)
flag=1;
else
continue;
if (mi>ma)//选出最小的最长时间
{
mi=ma;
p=i;
}
}
if (flag)
printf("%d %d\n",p,mi);
else
printf("disjoint\n");
}
return 0;
}
/*
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
*/