POJ 2263 Heavy Cargo 多种解法

好题。这题可以有三种解法：1.Dijkstra   2.优先队列   3.并查集

我这里是优先队列的实现，以后有时间再用另两种方法做做。。方法就是每次都选当前节点所连的权值最大的边，然后BFS搜索。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <functional>
using namespace std;
#define N 100007

struct node
{
    int ind,wt;
    bool operator < (const node &a)const
    {
        return wt<a.wt;
    }
};
int start,endi,n,m,res;
int way[][],vis[][];
map<string,int> mp;
priority_queue<node> que;

void BFS()
{
    memset(vis,,sizeof(vis));
    int i,maxi = ,id;
    node now,next;
    for(i=;i<=n;i++)
    {
        if(way[start][i] > maxi)
        {
            maxi = way[start][i];
            id = i;
        }
    }
    now.ind = id;
    now.wt = maxi;
    que.push(now);
    //printf("%d %d\n",start,endi);
    //printf("%d %d\n",now.ind,now.wt);
    res = ;
    while(!que.empty())
    {
        now = que.top();
        que.pop();
        if(now.ind == endi)
        {
            if(now.wt > res)
                res = now.wt;
            while(!que.empty())
                que.pop();
            return;
        }
        for(i=;i<=n;i++)
        {
            if(way[now.ind][i] && !vis[now.ind][i])
            {
                vis[now.ind][i] = vis[i][now.ind] = ;
                next.ind = i;
                next.wt = min(now.wt,way[now.ind][i]);
                que.push(next);
                //printf("%d %d\n",next.ind,next.wt);
            }
        }
    }
}

int main()
{
    int cs = ,i,j,w,num;
    string city1,city2;
    node ka,kb;
    while(scanf("%d%d",&n,&m)!=EOF && (n||m))
    {
        mp.clear();
        num = ;
        memset(way,,sizeof(way));
        for(i=;i<m;i++)
        {
            cin>>city1;
            cin>>city2;
            scanf("%d",&w);
            if(!mp[city1])
                mp[city1] = num++;
            if(!mp[city2])
                mp[city2] = num++;
            way[mp[city1]][mp[city2]] = w;
            way[mp[city2]][mp[city1]] = w;
        }
        cin>>city1>>city2;
        start = mp[city1];
        endi = mp[city2];
        BFS();
        printf("Scenario #%d\n",cs++);
        printf("%d tons\n\n",res);
    }
    return ;
}