poj 1125 最短路

2015/1/28

无脑的最短路，略烦。主要是题面有点醉（= =）。。。

每个点到其他点的最短路的长度Lmax,然后再去Lmax中去最小值。。

如果不存在，输出那个disjoint。

#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<climits>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>

using namespace std;
typedef long long ll;  

#define mod 10007
#define lson pos<<1,l,mid
#define sc(n) scanf("%d",&n)
#define rson pos<<1|1,mid+1,r
#define pr(n) printf("%d\n",n)
#define met(n,m) memset(n, m, sizeof(n))
#define F(x,y,i) for(int i = x;i > y; i--)
#define f(x,y,i) for(int i = x;i < y; i++)
#define ff(x,y,i) for(int i = x;i <= y; i++)
#define FF(x,y,i) for(int i = x;i >= y; i--) 

const int N=100500;
const int inf = INT_MAX/2;

int Max(int a,int b)
{
	return a>b?a:b;
}

int Min(int a,int b)
{
	return  a<b?a:b;
}

int s[1005][1005];
int dis[1005];
int tp[1005];

void init()
{
	f(0,1005,i)
	f(0,1005,j)s[i][j] = inf,s[i][i] = 0;
	met(tp,0);
}

int main()  
{  
    int n, m, tot, x, y;
    while(~scanf("%d",&n)&&n)
    {
    	init();
    	f(0,n,i)
    	{
	    	scanf("%d",&m);
	    	while(m--)
	    	{
	    		scanf("%d%d",&x,&y);
	    		s[i+1][x] = y;
	    	}
	    }
	    
	    f(1,n+1,k)
	    f(1,n+1,i)
	    f(1,n+1,j)
	    {
    		s[i][j] = Min(s[i][j],s[i][k]+s[k][j]);
    	}
    	f(1,n+1,i)
    	f(1,n+1,j)
    	{
    		if(i!=j)
	    	tp[i] = Max(tp[i],s[i][j]);
	    }
	    int minn = INT_MAX,point;
	    f(1,n+1,i)
	    {
    		if(minn > tp[i])
    		{
		    	point = i;
		    	minn = tp[i];
		    }
    	}
	   
	    
	    if(minn == INT_MAX)printf("disjoint\n");
	    else printf("%d %d\n",point,minn);
	    
    }
    return 0;  
}