题意:N个人,M条关系,A x y表示询问x和y是不是属于同一组,D x y表示x和y是不同组。输出每个询问后的结果。
分析:
1、所有的关系形成一个连通图,如果x和y可达,那两者关系是确定的,否则不能确定。
2、r[tmpy] = r[x] + r[y] + 1;可以更新连通块里祖先的标号。
eg:
5 4
D 1 2
D 2 3
D 4 5-----到此为止形成两个连通块,标号如图所示(红笔)
D 3 5
第四步,将3和5连边,因为以0为祖先,所以4的标号应当改变,可以发现改变后的r[4] = r[3] + r[5] + 1 = 2 + 1 + 1 = 4;
3、r[5]的更新在Find函数里完成。(Find函数更新原连通块里除祖先外的结点)
int tmp = Find(fa[x]);//更新5的原祖先4的权值
r[x] += r[fa[x]];//通过加上原祖先4的新权值以完成更新。
#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1e5 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int r[MAXN];
int fa[MAXN];
void init(){
for(int i = 0; i < MAXN; ++i){
fa[i] = i;
r[i] = 0;
}
}
int Find(int x){
if(fa[x] == x) return x;
int tmp = Find(fa[x]);
r[x] += r[fa[x]];
return fa[x] = tmp;
}
int main(){
int T;
scanf("%d", &T);
while(T--){
init();
int N, M;
scanf("%d%d", &N, &M);
for(int i = 0; i < M; ++i){
char a;
int x, y;
getchar();
scanf("%c%d%d", &a, &x, &y);
int tmpx = Find(x);
int tmpy = Find(y);
if(a == 'D'){
if(tmpx == tmpy) continue;
if(tmpx < tmpy){
fa[tmpy] = tmpx;
r[tmpy] = r[x] + r[y] + 1;
}
else{
fa[tmpx] = tmpy;
r[tmpx] = r[x] + r[y] + 1;
}
}
else if(a == 'A'){
if(tmpx != tmpy){
printf("Not sure yet.\n");
continue;
}
if(abs(r[y] - r[x]) & 1){
printf("In different gangs.\n");
}
else{
printf("In the same gang.\n");
}
}
}
}
return 0;
}