（POJ 1703）Find them, Catch them

Posted 2020-10-02 橘生淮南终洛枳

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 46763		Accepted: 14398

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

POJ Monthly--2004.07.18

题意：
　　有一群人属于两个队伍每行给出两个人说明他们不属于同一个队问给出两个人他们属于一个队还是不同的队或是不确定

思路：
　　并查集把给出的人分成几个集合每个集合之间的人的关系不确定，对同一个集合保存和本人不为同一队的人本着敌人的敌人便是朋友的原则用并查集同一集合为同一队不同集合为不同队

上代码：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

const int M = 200010;
int f[M],T,n,m,enemy[M];

int find(int x) {
    if(f[x]!=x) f[x]=find(f[x]);
    return f[x];
}

void merge(int a,int b) {
    int fa=find(a), fb=find(b);
    if(fa!=fb) 
        f[fb]=fa;
}

void work() {
    int a,b;
    scanf("%d%d",&n,&m);
    for(int i=1; i<=n; i++) f[i]=i;
    memset(enemy,0,sizeof(enemy));
    for(int i=1; i<=m; i++) {
        char type;
        getchar();
        scanf("%c%d%d",&type,&a,&b);
        if(type==‘D‘) {
            if(enemy[a]) merge(enemy[a],b);
            if(enemy[b]) merge(enemy[b],a);
            enemy[a]=b;
            enemy[b]=a;
        }
        else {
            if(find(a)==find(enemy[b])) printf("In different gangs.\n");
            else if(find(a)==find(b)) printf("In the same gang.\n");
            else printf("Not sure yet.\n"); 
        }
    }
    return ;
}

int main() {
    scanf("%d",&T);
    while(T--) work();
    return 0;
}

 

自己选的路，跪着也要走完！！！