hdu2544

一一题非常简单的单源无向正值最短路径图问题，用来练练手，分别用Dijkstra算法和prim算法做了一次，注释写得比较清楚了，看不懂可以先从下面的链接了解一下算法。

Dijkstra：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define loop(i,u,v) for(int i = u; i <= v; i++)
#define INF 0xffffff
#define N 105

int map[N][N],dis[N];
bool state[N];


void dijkstra(int n,int m)
{
    dis[1] = 0;
    state[1] = true;
    int cut = 1; //最新加入已经true的点
    loop(j,1,n)
    {
        if(state[j] == false)
        {
            if(dis[j] > dis[cut] + map[cut][j])
                dis[j] = dis[cut] + map[cut][j];
        }
    }//更新所有点到起点的距离
    loop(i,1,n)
    {
        int current_shortest = INF;
        int possible_point;
        loop(j,1,n)
        {
            if(state[j] == false)
            {
                if(current_shortest > dis[j] )
                {
                    current_shortest = dis[j] ;
                    possible_point = j;
                }
            }
        }//这个循环用来寻找集合[T]到集合[s]最近的点
        state[possible_point] = true;
        cut = possible_point;
        //cout << "selected point" << cut << endl;
        if(state[n] == true)
            break;
        loop(j,1,n)
        {
            if(state[j] == false)
            {
                if(dis[j] > dis[cut] + map[cut][j])
                    dis[j] = dis[cut] + map[cut][j];
            }
        }     //这个循环通过当前离起点最近的点来更新所有未并入集合[S]的点到起点的距离

    }
}

int main()
{
    int n,m;
    while(cin >> n >> m)
    {
        if(n == 0)
            break;
        memset(state,0,sizeof(state));
        loop(i,1,n)
            dis[i] = INF;
        loop(i,1,n)
        {
            loop(j,1,n)
                map[i][j] = INF;
        }
        int a,b,c;
        loop(i,1,m)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(c < map[a][b])
                map[a][b] = map[b][a] = c;
        }
        dijkstra(n,m);
        cout << dis[n] << endl;
    }

    return 0;
}

prim：

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

#define N 105
#define INF 0xffffff

int map[N][N];

void prim(int n)
{
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
            for(int k = 1; k <= n; k++)
            {
                map[j][i] = map[i][j] = min(map[i][j],map[i][k]+map[k][j]); 
            }
        }//DP的思想，就是通过k点，可以找到i->k->j这条路比i->j短
    }
}

int main()
{
    int n,m;
    while(cin >> n >> m)
    {
        if(n == 0)
            break;
        int a,b,c;
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                map[i][j] = INF;
            }
            map[i][i] = 0;//注意，一定要将这条路长度设为0
        }
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(map[a][b] > c)
                map[a][b] = map[b][a] = c;
        }
        prim(n);
        cout << map[1][n] << endl;
    }
    return 0;
}

http://www.cnblogs.com/biyeymyhjob/archive/2012/07/31/2615833.html
http://blog.youkuaiyun.com/zhongyanghu27/article/details/8221276