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POJ---1703 Find them, Catch them【并查集】

2017-04-24 20:41 447 查看
        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 identifywhich 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] 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.InputThe 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 lineseach containing one message as described above.OutputFor 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 "Notsure 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.
[b]Source：
题意：点击打开链接
城市中有两个帮派，D 1 2表示1与2不在同一个帮派，A 1 2表示查询1与2是否属于同一帮派
代码：
#include <iostream>#include<stdio.h>#include<string.h>#define maxn 200000using namespace std;int par[maxn];int enemy[maxn];//用来存放敌人，表示i与enemy[i]不在同一组int find(int a){    return a==par[a]?a:par[a]=find(par[a]);}int same(int a,int4000b){    return find(a) == find(b);}void unite(int a,int b){    a=find(a);    b=find(b);    if(a==b)        return;    par[a]=b;}int main(void){    int k,m,n,i,b,a;    char ch[2];    scanf("%d",&k);    while (k--)    {        memset(enemy,0,sizeof(enemy));        scanf("%d%d",&n,&m);        for(i = 1; i <= n; i++)            par[i] = i;        for(i = 0; i < m; i++)        {            scanf("%s%d%d",ch,&a,&b);//用字符数组，是为了处理回车符            if (*ch == 'D')            {                if (enemy[a])                    unite(enemy[a],b);//a的敌人便是b的朋友，将a的敌人与b联合起来                if (enemy[b])                    unite(enemy[b],a);//b的敌人便是a的朋友，将b的敌人与a联合起来                enemy[a] = b;//将a与b标记为敌人                enemy[b] = a;            }            else            {                if (find(a) == find(enemy[b]))//a与b的敌人属于同一组，则a与b便不属于同一组                    printf("In different gangs.\n");                else if (find(a) == find(b))                    printf("In the same gang.\n");                else                    printf("Not sure yet.\n");            }        }    }    return 0;}<span class="Apple-style-span" style="font-weight: bold;"></span>

                                            

                                        

                        
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