概述
The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs in the city, Gang Dragon and Gang Snake. However, the police first needs to identify which gang a criminal belongs to. The present question is, given two criminals; do they belong to a same clan? You must give your judgment based on incomplete information. (Since the gangsters are always acting secretly.)
Assume N (N <= 10^5) criminals are currently in Tadu City, numbered from 1 to N. And of course, at least one of them belongs to Gang Dragon, and the same for Gang Snake. You will be given M (M <= 10^5) messages in sequence, which are in the following two kinds:
D [a] [b]
where [a] and [b] are the numbers of two criminals, and they belong to different gangs.A [a] [b]
where [a] and [b] are the numbers of two criminals. This requires you to decide whether a and b belong to a same gang.
Input
The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case begins with a line with two integers N and M, followed by M lines each containing one message as described above.
Output
For each message “A [a] [b]” in each case, your program should give the judgment based on the information got before. The answers might be one of “In the same gang.”, “In different gangs.” and “Not sure yet.”
Sample Input
1
5 5
A 1 2
D 1 2
A 1 2
D 2 4
A 1 4
Sample Output
Not sure yet.
In different gangs.
In the same gang.
题意:第一行表示有多少测试用例。第二行n,m表示用多少个人以及下面有多少命令。其中如果是A则代表询问你A后面的这两人是不是一伙如果一伙则输出In the same gang..如果不是一伙则输出In different gangs.如果不确定输出Not sure yet。D代表的是D后面的两个人在不同的一伙。
思路:刚开始想用逻辑判断但是行不通。后来知道用并查集,但有一些技巧在里面。就是开一个数组将这个人的敌人放在里面。
AC代码:
include
include
using namespace std;
int fa[200005],enemy[200005];
int getfa(int a)
{
if (fa[a] == a)
return a;
else
fa[a] = getfa(fa[a]);
return fa[a];
}
void unite(int a,int b)
{
fa[getfa(b)] = getfa(a);
}
void work()
{
int m,n,i,b,a;
char ch;
memset(enemy,0,sizeof(enemy));
scanf(“%d%d”,&n,&m);
for(i = 1; i <= n; i++)
fa[i] = i;
for(i = 1; i <= m; i++)
{
getchar();
scanf(“%c%d%d”,&ch,&a,&b);
if (ch == ‘D’)
{
if(enemy[a])
unite(enemy[a],b);
if(enemy[b])
unite(enemy[b],a);
enemy[a] = b;
enemy[b] = a;
}
else
{
if (getfa(a) == getfa(enemy[b]))
printf(“In different gangs.n”);
else if (getfa(a) == getfa(b))
printf(“In the same gang.n”);
else
printf(“Not sure yet.n”);
}
}
return ;
}
int main()
{
int k;
scanf(“%d”,&k);
while (k–)
work();
return 0;
}
最后
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