POJ 3615 floyd 求任意起点终点的最短路

http://poj.org/problem?id=3615

题意：求起点到终点的最短路，不存在则输出-1.这题居然tle两次，把floyd放在外面就行了。

// I'm lanjiangzhou
//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

const int maxn =1010;
int edges[maxn][maxn];
int n,m,t;
int u,v,w;
int maxx(int a,int b){
    return a>b?a:b;
}
int minn(int a,int b){
    return a>b?b:a;
}

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

int main(){
    while(scanf("%d%d%d",&n,&m,&t)!=EOF)
    {
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                edges[i][j]=INF;
            }
            edges[i][i]=0;
        }
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&u,&v,&w);
            edges[u][v]=w;
        }
        int start,end;
        floyd();//放在外面，里面超时
        for(int i=1;i<=t;i++){
            scanf("%d%d",&start,&end);

            if(edges[start][end]==INF){
                printf("-1\n");
            }
            else printf("%d\n",edges[start][end]);
        }
    }
        //floyd();
    return 0;
}