2075:Tangled in Cables

总时间限制: 1000ms 内存限制: 65536kB
描述
You are the owner of SmallCableCo and have purchased the franchise rights for a small town. Unfortunately, you lack enough funds to start your business properly and are relying on parts you have found in an old warehouse you bought. Among your finds is a single spool of cable and a lot of connectors. You want to figure out whether you have enough cable to connect every house in town. You have a map of town with the distances for all the paths you may use to run your cable between the houses. You want to calculate the shortest length of cable you must have to connect all of the houses together.
输入
Only one town will be given in an input.
The first line gives the length of cable on the spool as a real number.
The second line contains the number of houses, N
The next N lines give the name of each house’s owner. Each name consists of up to 20 characters {a–z,A–Z,0–9} and contains no whitespace or punctuation.
Next line: M, number of paths between houses
next M lines in the form

< house name A > < house name B > < distance >
Where the two house names match two different names in the list above and the distance is a positive real number. There will not be two paths between the same pair of houses.
输出
The output will consist of a single line. If there is not enough cable to connect all of the houses in the town, output
Not enough cable
If there is enough cable, then output
Need < X > miles of cable
Print X to the nearest tenth of a mile (0.1).
样例输入
100.0
4
Jones
Smiths
Howards
Wangs
5
Jones Smiths 2.0
Jones Howards 4.2
Jones Wangs 6.7
Howards Wangs 4.0
Smiths Wangs 10.0
样例输出
Need 10.2 miles of cable

分析：

一定注意要初始化两个数组，path数组和cost数组。最小生成树步骤：先把数据用二维数组存储边的权重，然后调用prim，每次找到维护的cost数组最小值所代表顶点，然后更新cost数组，更新的时候不需要判断是否遍历过。

#include <iostream>#include<map>#include<iterator>#include<string>#include<algorithm>using namespace std;//http://bailian.openjudge.cn/practice/2075/ //很直接的最小生成树prim算法，注意的是这个是无向图 const int INF=100000000;int v,e;double path[510][510],m_l,res=0;//path是路径权重 double cost[510];//X集合到达其他顶点的最小值 bool used[510];map<string,int> m;map<string,int>::iterator it;void prim(){    for(int i=1;i<=v;i++){        cost[i]=INF;        used[i]=false;    }    cost[1]=0;    while(true){        int w=-1;//从不属于X 的集合中选取从X到其权值最小的顶点         for(int i=1;i<=v;i++){            if(used[i]==false&&(w==-1||cost[w]>cost[i])){                w=i;            }        }        if(w==-1)break;        used[w]=true;//把顶点放入X 集合         res+=cost[w];        for(int i=1;i<=v;i++){            cost[i]=min(cost[i],path[w][i]);            //更新 的时候没有if(used[i]==false)?         }    }}int main(int argc, char *argv[]) {    cin>>m_l>>v;    for(int i=1;i<=v;i++){        string tmp;        cin>>tmp;        m[tmp]=i;    }    for(int i=1;i<=v;i++){        for(int j=1;j<=v;j++){            path[i][j]=INF;//初始化权重都为INF         }    }    cin>>e;    for(int i=1;i<=e;i++){        string tmp1,tmp2;        double len;        cin>>tmp1>>tmp2>>len;        it=m.find(tmp1);        int u1=it->second;        it=m.find(tmp2);        int u2=it->second;        path[u1][u2]=len;//只要这一条语句的时候是有向图         path[u2][u1]=len;    }    prim();    if(res<=m_l)        cout<<"Need "<<res<<" miles of cable"<<endl;    else cout<<"Not enough cable"<<endl;     return 0;}