hdu 1874 最短路 Dijkstra Folyd

水题，唯一要注意的就是输入路径的时候注意判重。
先给出Dijkstra的代码：#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

int n, m;
int map[205][205];
int dis[205];
int vis[205];
const int INF = 50000000;
int sta, endd;

void Dijkstra( )
{
memset(vis, 0, sizeof(vis));
for(int i = 0; i < n; i++)
dis[i] = INF;
dis[sta] = 0;
for(int i = 0; i < n; i++)
{
int x, mm = INF;
for(int j = 0; j < n; j++)
if(!vis[j] && dis[j] <= mm)
{
x = j;
mm = dis[j];
}
vis[x] = 1;
for(int j = 0; j < n; j++)
if(dis[j] > dis[x] + map[x][j])
dis[j] = dis[x] + map[x][j];
}
}

int main()
{
// freopen("ztest.txt","r",stdin);
while(scanf("%d%d", &n, &m) == 2 )
{
for(int i = 0; i < 205; i++)
for(int j = 0; j < 205; j++)
map[i][j] = INF;
for(int i = 0; i < m; i++)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
if(c < map[a][b])  //不加这句就等着WA吧。。。。。
map[a][b] =map[b][a] = c;
}
scanf("%d%d", &sta, &endd);
Dijkstra( );
if(dis[endd] < INF )
printf("%d\n",dis[endd]);
else
printf("-1\n");
}
return 0;
}
给出Floyd算法的AC代码，简单，但是时间复杂度高，为O(n^3)。#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;

int d[205][205];
int n, m;
int sta, endd;
int t = 1;

const int INF = 50000000;

void Floyd( )
{
for(int i = 0; i < n; i++)
d[i][i] = 0;
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
for(int k = 0; k < n; k++)
d[j][k] = min(d[j][k], d[j][i] + d[i][k]);
}

int main()
{
//freopen("ztest.txt","r",stdin);
while(scanf("%d%d",&n, &m) == 2)
{
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
d[i][j] = INF;
for(int i = 0; i < m; i++)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
if(c < d[a][b])
d[a][b] = d[b][a] = c;
}
scanf("%d%d", &sta, &endd);
Floyd( );
if(d[sta][endd] < INF )
printf("%d\n",d[sta][endd]);
else
printf("-1\n");
}
return 0;
}