题意:
两个两个输入一些数,D代表这两个数不同,A代表查询这两个数是否相同
要点:
与食物链那一题类似,都是开多倍数组,储存每种情况
15321734 | Seasonal | 1703 | Accepted | 1680K | 375MS | C++ | 973B | 2016-03-27 18:34:06 |
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define maxn 100050
int p[maxn * 2], rank[maxn * 2];
int m, n;
void init()
{
for (int i = 1; i <= 2*m; ++i)//注意这里初始化要2*m
{
p[i] = i;
rank[i] = 0;
}
}
int find(int x)
{
if (p[x] == x) return x;
return p[x] = find(p[x]);
}
void merge(int x, int y)
{
x = find(x);
y = find(y);
if (x == y) return;
if (rank[x] > rank[y])
p[y] = x;
else
{
p[x] = y;
if (rank[x] == rank[y]) ++rank[y];
}
}
bool same(int x, int y)
{
return find(x) == find(y);
}
int main()
{
int t,i,x,y;
char str;
scanf("%d", &t);
while (t--)
{
scanf("%d%d", &m, &n);
getchar();
init();
while(n--)
{
scanf("%c%d%d", &str,&x,&y);
getchar(); //这一步比较重要,清除缓冲区
if (str == 'A')
{
if (same(x, y))
printf("In the same gang.\n");
else if (same(x, y + m))
printf("In different gangs.\n");
else
printf("Not sure yet.\n");
}
else
{
merge(x, y + m);//合并X为A,Y为B的情况
merge(x + m, y);//合并X为B,Y为A的情况
}
}
}
return 0;
}