并查集找gangs

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题目大意是有两个gangs,每次给你两个来自不同gangs的人,然后再问你任两个人的gangs关系。

解题思路是:将数据的最大范围扩大两倍,a 与 b 来自不同的gans,则a 与 b +n 来自相同的gangs,b 与 a+n来自相同的gangs.

判断相同gangs的条件是:findfather(a) == findfather(b);

判断不相同gangs的条件是: findfather(a) == findfather(b+n) || findfather(b) == findfather(a+n);

不确定关系的条件是:findfather(a) != findfather(b) && findfather(a)!=findfather(b+n) ;

代码如下:

#include <iostream>#include <cstdio>using namespace std;const int maxn = 100000+100;int Set[2 * maxn] ;int n , m ;int t , fx , fy;void init(){    for(int i = 0; i < 2 * maxn ; i ++){        Set[i] = i ;    }}int findfather(int x){   if(x == Set[x]) return x ;   return Set[x] = findfather(Set[x]) ;}void union_ab(int x , int y){    int fx = findfather(x) ;    int fy = findfather(y) ;    if(fx != fy) Set[fx] = fy ;}int main(){    int a , b ;    char c[3] ;    scanf("%d",&t);    while(t --){        init();        scanf("%d%d",&n , &m);        for(int i = 0; i < m; i ++){            scanf("%s%d%d",c,&a,&b);            if(c[0] =='D') {                union_ab(a , b + n);                union_ab(a + n , b);            }            else            {                if(findfather(a) == findfather(b))                     printf("In the same gang.\n");                else if(findfather(a) == findfather(b+n) || findfather(a) == findfather(b))                          printf("In different gangs.\n");                   else                        if(findfather(a+n)!=findfather(b) && findfather(b) != findfather(a))                               printf("Not sure yet.\n");            }        }    }    return 0;}
方法二:
#include <iostream>#include <cstdio>using namespace std;const int maxn = 100005;typedef struct{   int root, relation;}gang;gang data[maxn];int t , n , m ;void init(){    for(int i = 0; i < maxn; i ++)    {        data[i].root = i;        data[i].relation = 0;    }}int findfather(int x){    if(x == data[x].root) return x;    int t = findfather(data[x].root);    data[x].relation = (data[x].relation + data[data[x].root].relation)%2;    data[x].root = t;    return t;}void union_ab(int x , int y ){    int fx = findfather(x);    int fy = findfather(y);    if(fx != fy){        if(data[x].relation == data[y].relation)        {            data[fx].relation = 1;            data[fx].root = fy;        }        if(data[x].relation != data[y].relation)        {            data[fx].root = fy;        }    }}int main(){    scanf("%d",&t);    char c[10];    int a , b;    while(t --)    {        init();        scanf("%d %d",&n ,&m);        for(int i = 0; i < m; i ++)        {            scanf("%s",c);            if(c[0] == 'D')            {                scanf("%d%d",&a,&b);                union_ab(a , b);            }            if(c[0] == 'A'){                scanf("%d%d",&a , &b);                findfather(a);                findfather(b);                if(data[a].root != data[b].root)                   printf("Not sure yet.\n");                else                {                    if(data[a].relation == data[b].relation)                                 printf("In the same gang.\n");                      else if(data[a].relation != data[b].relation)                              printf("In different gangs.\n");                }            }        }    }    return 0;}


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