标签：uva
| Network Connections |
Bob, who is a network administrator, supervises a network of computers. He is keeping a log connections between the computers in the network. Each connection is bi-directional. Two computers are interconnected if they are directly connected or if they are interconnected with the same computer. Occasionally, Bob has to decide, quickly, whether two given computers are connected, directly or indirectly, according to the log information.
Write a program which based on information input from a text file counts the number of successful and the number of unsuccessful answers to the questions of the kind :
is computeri interconnected with computerj ?
. A pair of this form shows that computeri and computerj get
interconnected.. A pair of this form stands for
the question: is computeri interconnectedwith computerj?There‘s a blank line between datasets.
Each pair is on a separate line. Pairs can appear in any order, regardless of their type. The log is updated after each pair of type (a) and each pair of type (b) is processed according to the current network configuration.
For example, the input file illustrated in the sample below corresponds to a network of 10 computers and 7 pairs. There are N1 successfully answered questions and N2 unsuccessfully answered questions. The program
prints these two numbers to the standard output on the same line, in the order: successful answers, unsuccessful answers, as shown in the sample output. Print a blank line between datasets.
1 10 c 1 5 c 2 7 q 7 1 c 3 9 q 9 6 c 2 5 q 7 5
1,2
每次q后计算失败和成功的次数，最后输出成功的和失败的次数。
解题思路：
并查集，死了很多次在输入上面，输入还是需要一些技巧的。
代码：
#include<iostream>
#include<cstdio>
#include<string>
using namespace std;
const int maxn=1000000;//之前一直WA忘记了str0也可能随数字的增大而增长，
int t,n,parent[maxn],success,fail;
char str0[100];
void initial(){
for(int i=0;i<=n;i++){parent[i]=i;}
success=fail=0;
}
void outPut(){
printf("%d,%d\n",success,fail);
if(t) printf("\n");
}
int getRoot(int k){
if(k!=parent[k]) parent[k]=getRoot(parent[k]);
return parent[k];
}
void Union(int a,int b){
parent[a]=b;
}
void solve(){
int p,q,a,b;
sscanf(str0+1,"%d%d",&a,&b);//从字符串里面截取数字的函数，以空格分数字.
p=getRoot(a);
q=getRoot(b);
if(str0[0]=='c'){if(p!=q) Union(p,q);}
if(str0[0]=='q'){
if(p==q) success++;
else fail++;
}
}
int main(){
scanf("%d",&t);
while(t--){
cin>>n;
getchar();
initial();
while(gets(str0)){
if(!str0[0]) break;//判断是否为空行
solve();
}
outPut();
}
return 0;
}
Network Connections UVA 793（并查集）
标签：uva
原文地址：http://blog.csdn.net/hush_lei/article/details/41172585
