hdu2544 （bellman-ford)

题意：求顶点 1 - 顶点n 之间的最短路

解题思路：bellman-ford 算法，利用n-1松弛操作得到最短路，算法复杂度为 V*E

解题代码：

// File Name: 2544_1.cpp
// Author: darkdream
// Created Time: 2014年04月04日 星期五 22时37分45秒

#include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<sstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<climits>

using namespace std;

struct node{
    int x, d ; 
    node()
    {}
    node(int p,int q)
    {
        x = p; 
        d = q; 
    }
};
vector <node> edge[200];
int n , m ; 
int dis[200];
int bellman_ford(int s)
{
    for(int i = 1;i <= n;i ++) dis[i] = 1 << 27; 
    dis[s] = 0 ; 
    for(int i = 1;i <= n -1;i ++)
    {
        for(int j = 1;j <= n;j ++)
        {
            int k = edge[j].size();
            for(int p = 0 ;p < k ;p ++)
            {
                if(dis[edge[j][p].x] > dis[j] + edge[j][p].d)
                    dis[edge[j][p].x] = dis[j] + edge[j][p].d; 
            }
        }
    }
    return 1;   
}
int main(){
    //freopen("/home/plac/problem/input.txt","r",stdin);
    //freopen("/home/plac/problem/output.txt","w",stdout);
    while(scanf("%d %d",&n,&m)!= EOF)
    {
        if(n == 0 && m == 0 )
            break;
        for(int i =1 ;i <= n;i++)
            edge[i].clear();
        for(int i = 1;i <= m ;i ++)
        { 
            int a, b, c;
            scanf("%d %d %d",&a,&b,&c);
            edge[a].push_back(node(b,c));
            edge[b].push_back(node(a,c));
        }
        if(bellman_ford(1))
            printf("%d
",dis[n]);
    }
    return 0;
}