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[最新] Find them, Catch them(poj 1703)

qq41061455 21天前 0

Find them, Catch them

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 131   Accepted Submission(s) : 22

Problem Description

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: 
1. D [a] [b] 
where [a] and [b] are the numbers of two criminals, and they belong to different gangs. 
2. 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.

Source

PKU

对于每个" D a b "操作,union(a,b),这样做的目的是让a,b同根,方便判断是否帮派已确定,以及通过re[a],re[b] 确认是否在同一帮派中,其中re[a]表示与其根节点的关系(0 同,1异),如果re[a]==re[b] 那么在同一帮派,否则不在。所以关键就是更新re数组

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int fa[maxn],re[maxn];
char s[3];
int root(int x)
{
    if(x==fa[x]) return x;
    int t=fa[x];
    fa[x]=root(fa[x]);
    re[x]=(re[x]+re[t])%2;
    return fa[x];
}
void make_re(int x,int y)
{
    int fx=root(x);
    int fy=root(y);
    fa[fx]=fy;
    re[fx]=(re[x]+1+re[y])%2;
}
int main()
{
    int t,n,m,a,b;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++) fa[i]=i,re[i]=0;
        while(m--)
        {
            scanf("%s%d%d",s,&a,&b);
            if(s[0]=='A')
            {
                if(root(a)==root(b))
                {
                    if(re[a]!=re[b]) printf("In different gangs.\n");
                    else printf("In the same gang.\n");
                }
                else printf("Not sure yet.\n");
            }
            else make_re(a,b);
        }
    }
}

 

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