POJ 2263 Heavy Cargo 多种解法

来源：互联网发布：软件开发方法学编辑：程序博客网时间：2024/04/29 15:52

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 100007struct node{    int ind,wt;    bool operator < (const node &a)const    {        return wt<a.wt;    }};int start,endi,n,m,res;int way[203][203],vis[203][203];map<string,int> mp;priority_queue<node> que;void BFS(){    memset(vis,0,sizeof(vis));    int i,maxi = 0,id;    node now,next;    for(i=1;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 = 0;    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=1;i<=n;i++)        {            if(way[now.ind][i] && !vis[now.ind][i])            {                vis[now.ind][i] = vis[i][now.ind] = 1;                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 = 1,i,j,w,num;    string city1,city2;    node ka,kb;    while(scanf("%d%d",&n,&m)!=EOF && (n||m))    {        mp.clear();        num = 1;        memset(way,0,sizeof(way));        for(i=0;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 0;}

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