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。

View Code

// 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;   //codeforcestypedef unsigned int ui;typedef unsigned long long ui64;#define MP make_pairtypedef 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")//endconst 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;}