学步园

事实证明 ，我只能切水题， floyd的循环抽时间得搞明白啊啊啊

/*
ID : Your ID
LANG : C++
PROB : namenum
*/
#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <limits>
#include <vector>
#include <bitset>
#include <string>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <string.h>
#include <iostream>
#include <algorithm>
#define Si set<int>
#define LL long long
#define pb push_back
#define PS printf(" ")
#define Vi vector<int>
#define LN printf("\n")
#define lson l,m,rt << 1
#define rson m+1,r,rt<<1|1
#define SD(a) scanf("%d",&a)
#define PD(a) printf("%d",a)
#define SET(a,b) memset(a,b,sizeof(a))
#define FF(i,a) for(int i(0);i<(a);i++)
#define FD(i,a) for(int i(a);i>=(1);i--)
#define FOR(i,a,b) for(int i(a);i<=(b);i++)
#define FOD(i,a,b) for(int i(a);i>=(b);i--)
#define readf freopen("input.txt","r",stdin)
#define writef freopen("output.txt","w",stdout)
const int maxn = 1005;
const int INF = 0x3fffffff;
const int dx[]={0,1,0,-1};
const int dy[]={1,0,-1,0};
const double pi = acos(-1.0);
using namespace std;

int dis[101][101];

int N;

void floyd()
{
	FOR(k,1,N) FOR(i,1,N) FOR(j,1,N){
        if(i!=j && dis[i][j] > dis[i][k] + dis[k][j])
            dis[i][j] = dis[i][k] + dis[k][j];
	}
	int maxlength;
	int min_in_max=INF;
	int flag_source;

	FOR(i,1,N){
		maxlength=0;
		FOR(j,1,N)
			if(i!=j && maxlength<dis[i][j])   
				maxlength=dis[i][j];
		if(min_in_max>maxlength){
			min_in_max=maxlength;      
			flag_source=i;             
		}
	}

	if(min_in_max<INF)
		cout<<flag_source<<' '<<min_in_max<<endl;
	else
		cout<<"disjoint"<<endl;
	return;
}

int main(){

	while(~SD(N) && N){
		FOR(i,1,N) FOR(j,1,N) dis[i][j]=INF;
		FOR(i,1,N){
			int pair;
			SD(pair);
            FOR(j,1,pair){
				int cat,time;
				scanf("%d%d",&cat,&time);
				dis[i][cat]=time;
			}
		}

		floyd();
	}
	return 0;
}