Tangled in Cables


Time Limit: 2 Seconds      Memory Limit: 65536 KB


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, N
  • The next N lines give the name of each house's owner. Each name consists of up to 20 characters {a�Cz,A�CZ,0�C9} 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.

Output

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).

Sample Input

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

Sample Output

Need 10.2 miles of cable

一道奇妙的最小生成树。最终变了花样。一般easy出现
上述情况,就是数组开的小了.
可是在zoj过了,在杭电wa了。真是一道奇妙的题;
附ac代码:
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
int start;
int end;
double cost;
}t[1001000];
int cmp(node a,node b)
{
return a.cost <b.cost ;
}
int per[11000];
int find(int x)
{
int r=x;
while(r!=per[r])
r=per[r];
return r;
}
int join(int x,int y)
{
int fx=find(x);
int fy=find(y);
if(fx!=fy)
{
per[fx]=fy;
return 1;
}
return 0;
}
int main()
{
double total;
int i,n,m,j,k;
char c[10000][25],s[25],ch[25];
scanf("%lf",&total);
for(i=0;i<11000;i++)
per[i]=i;
scanf("%d",&m);
for(i=0;i<m;i++)
scanf("%s",c[i]);
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%s%s%lf",s,ch,&t[i].cost);
for(j=0;j<m;j++)
{
if(strcmp(s,c[j])==0)
t[i].start=j;
if(strcmp(ch,c[j])==0)
t[i].end =j;
}
}
sort(t,t+n,cmp);
double sum=0.0;
for(i=0;i<n;i++)
if(join(t[i].start,t[i].end))
sum=sum+t[i].cost;
if(sum>total)
printf("Not enough cable\n");
else
printf("Need %.1lf miles of cable\n",sum);
return 0;
}

用prim在杭电能过,附杭电ac代码:

<pre name="code" class="cpp">#include<stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#define N 10000
#define M 1000000
#include <algorithm>
using namespace std;
int n, m, ncount, pre[N];
char name[N][21];
float sum_len, min_len;
struct edge
{
int x, y;
float len;
}node[M]; bool cmp(struct edge a, struct edge b)
{
return a.len < b.len;
} int search_name(char * s)
{
int i;
for(i=1; i<=n; i++)
if( !strcmp(s, name[i]) )
return i;
} void input()
{
int i, x, y;
char s[21];
scanf("%d", &n);
for(i=1; i<=n; i++)
scanf("%s", name[i]);
scanf("%d", &m);
for(i=1; i<=m; i++)
{
scanf("%s", s);
x = search_name(s);
scanf("%s", s);
y = search_name(s);
ncount++;
node[ ncount ].x = x;
node[ ncount ].y = y;
scanf("%f", &node[ ncount ].len);
}
}
int find_pre(int x)
{
while( x != pre[x] )
x = pre[x];
return x;
}
void kruskal()
{
int i, j, a, b;
for(i=1; i<=n; i++)
pre[i] = i;
sort(node+1, node+m+1, cmp);
for(i=1; i<=m; i++)
{
a = find_pre( node[i].x );
b = find_pre( node[i].y );
if( a != b )
{
min_len += node[i].len;
pre[ b ] = a;
}
}
if(sum_len < min_len)
printf("Not enough cable\n");
else
printf("Need %.1f miles of cable\n", min_len);
} int main()
{
int i;
while( scanf("%f", &sum_len) != EOF )
{
ncount = 0;
min_len = 0;
input();
kruskal();
}
return 0;
}


ZOJ2326Tangled in Cables(最小生成树)的更多相关文章

  1. UVALive 4872 Underground Cables 最小生成树

    题目链接: 题目 Underground Cables Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %ll ...

  2. POJ 2075 Tangled in Cables 最小生成树

    简单的最小生成树,不过中间却弄了很久,究其原因,主要是第一次做生成树,很多细节不够熟练,find()函数的循环for判断条件是 pre[i]>=0,也就是遇到pre[i]==-1时停止,i就是并 ...

  3. 图论常用算法之一 POJ图论题集【转载】

    POJ图论分类[转] 一个很不错的图论分类,非常感谢原版的作者!!!在这里分享给大家,爱好图论的ACMer不寂寞了... (很抱歉没有找到此题集整理的原创作者,感谢知情的朋友给个原创链接) POJ:h ...

  4. poj2075

    Tangled in Cables Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6348   Accepted: 2505 ...

  5. UvaLive 4872 Underground Cables (最小生成树)

    题意: 就是裸的最小生成树(MST), 完全图, 边长是实数. 分析: 算是复习一下MST把 方法一: prim 复杂度(n^2) #include <bits/stdc++.h> usi ...

  6. URAL 1160 Network(最小生成树)

    Network Time limit: 1.0 secondMemory limit: 64 MB Andrew is working as system administrator and is p ...

  7. POJ 1287 Networking (最小生成树)

    Networking 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/B Description You are assigned ...

  8. ZOJ 1542 POJ 1861 Network 网络 最小生成树,求最长边,Kruskal算法

    题目连接:problemId=542" target="_blank">ZOJ 1542 POJ 1861 Network 网络 Network Time Limi ...

  9. POJ 1287 Networking(最小生成树)

    题意  给你n个点 m条边  求最小生成树的权 这是最裸的最小生成树了 #include<cstdio> #include<cstring> #include<algor ...

随机推荐

  1. js基础——事件绑定(事件监听)

    JavaScript事件一共有三种监听方法分别如下: 1.事件监听一夹杂在html标签内 <div id="box" onClick="alert('HELLO W ...

  2. shell 键盘录入和运算

    一.read 命令,从键盘读入数据,赋给变量 1.脚本代码 #!/bin/sh read arg1 arg2 echo "第一个参数: $arg1" echo "第二个参 ...

  3. WinServer-IIS-js无法加载问题

    IIS中无法加载JS文件错误 尝试下面的几种解决方法,一起用

  4. ASP.NET-RedirectToAction只能使用get方法

    两个同名Action共同使用return View() return RedirectToAction("test", new { ls = list.Fct_OrderList ...

  5. WinServer-IIS-URL重写

    WEB平台安装程序在Windows Server里面才有,在WIN7里面是没有的 然后在安装一个URL重写工具 然后再设置各种规则 来自为知笔记(Wiz)

  6. 不用while if 循环求 1到n的和,还用到了!!这样的运算符

    很好的题目.开始我也没有想出来. 不用while if 循环求 1到n的和 给了很多种解法,第一种是构造函数,然后累加static变量 第二种是虚函数,有父类子类,父类的结束,然后用 !!n来让 n不 ...

  7. hdoj-1312-Red and Black

    Red and Black Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total ...

  8. OSGI项目中获取文件路径

    假设想依据给定的文件名创建一个File实例,你可能会这么写: File file = new File(当前类.class.getResource("config").toURI( ...

  9. Android中Handler原理

    Handler主要是主线程和子线程通信.一般子线程中做一些耗时操作做完之后通知主线程来改动UI. 实际上android系统在Activity启动或者状态变化等都是通过Handler机制实现的. 首先进 ...

  10. MongoDB(一)——简介

    这两天简单学习了一下MongoDB数据库,属于NoSQL类型数据库的一种,先简单宏观的看一下NoSQL的相关知识和MongoDB的基础知识. NoSQL是Not Only SQL的缩写,它指的是非关系 ...