ZOJ 1082 floyd

题意:求从经济人散布谣言到所有人需要的最短时间的最大值

具体详解看我原来的解法http://blog.sina.com.cn/s/blog_99ca2df50101807p.html

// I'm the Topcoder
//C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
//C++
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cctype>
#include <stack>
#include <string>
#include <list>
#include <queue>
#include <map>
#include <vector>
#include <deque>
#include <set>
using namespace std;

//*************************OUTPUT*************************
#ifdef WIN32
#define INT64 "%I64d"
#define UINT64 "%I64u"
#else
#define INT64 "%lld"
#define UINT64 "%llu"
#endif

//**************************CONSTANT***********************
#define INF 0x3f3f3f3f
#define eps 1e-8
#define PI acos(-1.)
#define PI2 asin (1.);
typedef long long LL;
//typedef __int64 LL;   //codeforces
typedef unsigned int ui;
typedef unsigned long long ui64;
#define MP make_pair
typedef vector<int> VI;
typedef pair<int, int> PII;
#define pb push_back
#define mp make_pair

//***************************SENTENCE************************
#define CL(a,b) memset (a, b, sizeof (a))
#define sqr(a,b) sqrt ((double)(a)*(a) + (double)(b)*(b))
#define sqr3(a,b,c) sqrt((double)(a)*(a) + (double)(b)*(b) + (double)(c)*(c))

//****************************FUNCTION************************
template <typename T> double DIS(T va, T vb) { return sqr(va.x - vb.x, va.y - vb.y); }
template <class T> inline T INTEGER_LEN(T v) { int len = 1; while (v /= 10) ++len; return len; }
template <typename T> inline T square(T va, T vb) { return va * va + vb * vb; }

// aply for the memory of the stack
//#pragma comment (linker, "/STACK:1024000000,1024000000")
//end

int n;
const int maxn = 100+10;
int dis[maxn][maxn];
int cost[maxn];
int maxx=-INF,minx=INF;

void floyd(){
    for(int k=1;k<=n;k++){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
            }
        }
    }
}

int main(){
    while(scanf("%d",&n)!=EOF){
        if(n==0 ) break;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                dis[i][j]=INF;
            }
            dis[i][i]=0;
        }
        int m;
        int to,w;
        for(int i=1;i<=n;i++){
            scanf("%d",&m);
            for(int j=1;j<=m;j++){
                scanf("%d%d",&to,&w);
                dis[i][to]=w;
            }
        }
        floyd();
        int pos;
        minx=INF;
        for(int i=1;i<=n;i++){
            maxx=-INF;
            for(int j=1;j<=n;j++){
                if(dis[i][j]>maxx){
                    //dis[i][j]=maxx;
                    maxx=dis[i][j];
                }
            }
            if(maxx<minx){
                pos=i;
                minx=maxx;
            }
        }
        if(minx==INF) printf("disjoint\n");
        else printf("%d %d\n",pos,minx);

    }
    return 0;
}