Tangled in Cables
Time Limit: 1000MSMemory Limit: 30000K
Total Submissions: 6039Accepted: 2386
Description
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.
Input
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, NThe 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 housesnext 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.
Output
The output will consist of a single line. If there is not enough cable to connect all of the houses in the town, outputNot enough cableIf there is enough cable, then outputNeed < X > miles of cablePrint X to the nearest tenth of a mile (0.1).
Sample Input
100.04JonesSmithsHowardsWangs5Jones Smiths 2.0Jones Howards 4.2Jones Wangs 6.7Howards Wangs 4.0Smiths Wangs 10.0
Sample Output
Need 10.2 miles of cable
Source
题目链接：poj.org/problem?id=2075题目大意：给一个总线长，和n个人，，m组关系，表示两个人之间的距离，现在要求所有人之间都连通，问最少需要多长的线，若超过总长，则输出Not enough cable题目分析：裸最小生成树问题，直接map一下字符串，再跑一下Kruskal#include <cstdio>#include <cstring>#include <map>#include <algorithm>#include <iostream>using namespace std;int const MAX = 1e4;int fa[MAX];int re[MAX];int n, m;map<string, int> mp;struct Edge{int u, v;double w;}e[MAX];bool cmp(Edge a, Edge b){return a.w < b.w;}void UF_set(){for(int i = 0; i < MAX; i++)fa[i] = i;}int Find(int x){return x == fa[x] ? x : fa[x] = Find(fa[x]);}void Union(int a, int b){int r1 = Find(a);int r2 = Find(b);if(r1 != r2)fa[r2] =r1;}double Kruskal(){UF_set();int num = 0;double res = 0;for(int i = 0; i < m; i++){int u = e[i].u;int v = e[i].v;if(Find(u) != Find(v)){Union(u, v);res += e[i].w;num ++;}if(num >= n – 1)break;}return res;}int main(){string s, s1, s2;double val, sum, ans = 0;scanf("%lf", &sum);int cnt = 1;mp.clear();scanf("%d", &n);for(int i = 0; i < n; i++){cin >> s;if(!mp[s])mp[s] = cnt ++;}scanf("%d", &m);for(int i = 0; i < m; i++){cin >> s1 >> s2 >> val;e[i].u = mp[s1];e[i].v = mp[s2];e[i].w = val;}sort(e, e + m, cmp);ans = Kruskal();if(ans < sum)printf("Need %.1f miles of cable\n", ans);elseprintf("Not enough cable\n");}
如果可以，我真想和你一直旅行。
