2075:Tangled in Cables

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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

来源

Mid-Atlantic 2004

解析：题目大意是给定一些人和他们的家间需要的cable，问是否有足够的cable连接所有的house，把house当作点，需要的cable为边上的权值，求最小生成树即可。

代码：

#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<algorithm>
#include<vector>
#include<cstdio>
using namespace std;
map<string,int> gp;
struct Edge{int from,to;double cost;Edge(int f,int t,double c):from(f),to(t),cost(c){}bool operator < (const Edge & o)const{return cost < o.cost;}
};
int fa[1002];
int find(int x){if(fa[x] == -1) return x;return fa[x] = find(fa[x]);
}
vector<Edge> graph;int main(){double cable,co,total = 0; int n,m,eNum = 0; string str,str1;scanf("%lf",&cable);scanf("%d",&n);for(int i = 0; i < n; ++i){cin>>str;gp[str] = i;}cin>>m;for(int i = 0; i < m;i++){cin>>str>>str1>>co;graph.push_back(Edge(gp[str],gp[str1],co));}sort(graph.begin(),graph.end());memset(fa,-1,sizeof(fa));for(size_t i = 0; i < graph.size();i++){Edge e = graph[i];if(find(e.from) != find(e.to)){total += e.cost;eNum++;fa[find(e.from)] = find(e.to);if(eNum == n - 1) break;}}if(total > cable)puts("Not enough cable");else printf("Need %.01lf miles of cable\n",total);return 0;
}