POJ 1703 Find them, Catch them (并查集)
2017-03-06 20:54
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题意:一共有N个人,给出M个操作分为两种:
1、A a b :提问a和b是否是同一个帮派的。有三种答案:是,不是和不确定
2、D a b :a和b不是同一个帮派的。
析:加权并查集,用一个r[i]来表示 i 和其父亲的关系,如果为0,表示 i 和其父亲是同一帮,1表示不是,每次更新即可。
代码如下:
1、A a b :提问a和b是否是同一个帮派的。有三种答案:是,不是和不确定
2、D a b :a和b不是同一个帮派的。
析:加权并查集,用一个r[i]来表示 i 和其父亲的关系,如果为0,表示 i 和其父亲是同一帮,1表示不是,每次更新即可。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<double, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e5 + 10; const int mod = 1e6; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int p[maxn], r[maxn]; int Find(int x){ if(x == p[x]) return x; int tmp = p[x]; p[x] = Find(p[x]); r[x] = (r[x] + r[tmp]) % 2; return p[x]; } int main(){ int T; cin >> T; while(T--){ scanf("%d %d", &n, &m); for(int i = 1; i <= n; ++i) p[i] = i, r[i] = 0; char s[5]; while(m--){ int a, b; scanf("%s %d %d", s, &a, &b); if(s[0] == 'A'){ int x = Find(a); int y = Find(b); if(x != y) puts("Not sure yet."); else if(r[a] != r[b]) puts("In different gangs."); else puts("In the same gang."); } else { int x = Find(a); int y = Find(b); if(x == y) continue; p[y] = x; r[y] = (r[a] + r[b] + 1) % 2; } } } return 0; }
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