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Bob has N (1 ≤ N ≤ 2*105) gears (numbered from 1 to N). Each gear can rotate clockwise or counterclockwise. Bob thinks that assembling gears is much more exciting than just playing with a single one. Bob wants to put some gears into some groups. In each gear group, each gear has a specific rotation respectively, clockwise or counterclockwise, and as we all know, two gears can link together if and only if their rotations are different. At the beginning, each gear itself is a gear group.
Bob has M (1 ≤ N ≤ 4*105) operations to his gears group:
Since there are so many gears, Bob needs your help.
Input will consist of multiple test cases. In each case, the first line consists of two integers N and M. Following M lines, each line consists of one of the operations which are described above. Please process to the end of input.
For each query operation, you should output a line consist of the result.
3 7 L 1 2 L 2 3 Q 1 3 Q 2 3 D 2 Q 1 3 Q 2 3 5 10 L 1 2 L 2 3 L 4 5 Q 1 2 Q 1 3 Q 1 4 S 1 D 2 Q 2 3 S 1
Same Different Same Unknown Different Same Unknown 3 Unknown 2
Link (1, 2), (2, 3), (4, 5), gear 1 and gear 2 have different rotations, and gear 2 and gear 3 have different rotations, so we can know gear 1 and gear 3 have the same rotation, and we didn‘t link group (1, 2, 3) and group (4, 5), we don‘t know the situation about gear 1 and gear 4. Gear 1 is in the group (1, 2, 3), which has 3 gears. After putting gear 2 away, it may have a new rotation, and the group becomes (1, 3).
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib> 10 #include <string> 11 #include <set> 12 #define LL long long 13 #define INF 0x3f3f3f3f 14 using namespace std; 15 const int maxn = 1010000; 16 int fa[maxn],dis[maxn],sum[maxn],mp[maxn],t,n, m; 17 void init() { 18 for(int i = 0; i <= n+m; i++) { 19 fa[i] = i; 20 mp[i] = i; 21 dis[i] = 0; 22 sum[i] = 1; 23 } 24 t = n+1; 25 } 26 int Find(int x) { 27 if(fa[x] != x) { 28 int root = Find(fa[x]); 29 dis[x] += dis[fa[x]]; 30 fa[x] = root; 31 } 32 return fa[x]; 33 } 34 int main() { 35 char st[10]; 36 while(~scanf("%d %d",&n, &m)) { 37 init(); 38 int x, y; 39 for(int i = 0; i < m; i++) { 40 scanf("%s",st); 41 if(st[0] == ‘L‘) { 42 scanf("%d %d",&x, &y); 43 x = mp[x]; 44 y = mp[y]; 45 int tx = Find(x); 46 int ty = Find(y); 47 if(tx != ty) { 48 sum[tx] += sum[ty]; 49 fa[ty] = tx; 50 dis[ty] = dis[x]+dis[y]+1; 51 } 52 } else if(st[0] == ‘Q‘) { 53 scanf("%d %d",&x, &y); 54 x = mp[x]; 55 y = mp[y]; 56 if(Find(x) != Find(y)) puts("Unknown"); 57 else { 58 if(abs(dis[x]-dis[y])&1) puts("Different"); 59 else puts("Same"); 60 } 61 } else if(st[0] == ‘D‘) { 62 scanf("%d",&x); 63 int tx = mp[x]; 64 tx = Find(tx); 65 sum[tx] -= 1; 66 mp[x] = ++t; 67 } else if(st[0] == ‘S‘) { 68 scanf("%d",&x); 69 x = mp[x]; 70 int tx = Find(x); 71 printf("%d\n",sum[tx]); 72 } 73 } 74 } 75 return 0; 76 }
xtu read problem training 3 B - Gears,布布扣,bubuko.com
xtu read problem training 3 B - Gears
标签:des style blog http color java os io
原文地址:http://www.cnblogs.com/crackpotisback/p/3891866.html