1512 Monkey King
Monkey King
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6667 Accepted Submission(s):
2858
Problem Description
Once in a forest, there lived N aggressive monkeys. At
the beginning, they each does things in its own way and none of them knows each
other. But monkeys can't avoid quarrelling, and it only happens between two
monkeys who does not know each other. And when it happens, both the two monkeys
will invite the strongest friend of them, and duel. Of course, after the duel,
the two monkeys and all of there friends knows each other, and the quarrel above
will no longer happens between these monkeys even if they have ever
conflicted.
Assume that every money has a strongness value, which will be reduced to only half of the original after a duel(that is, 10 will be reduced to 5 and 5 will be reduced to 2).
And we also assume that every monkey knows himself. That is, when he is the strongest one in all of his friends, he himself will go to duel.
Assume that every money has a strongness value, which will be reduced to only half of the original after a duel(that is, 10 will be reduced to 5 and 5 will be reduced to 2).
And we also assume that every monkey knows himself. That is, when he is the strongest one in all of his friends, he himself will go to duel.
Input
There are several test cases, and each case consists of
two parts.
First part: The first line contains an integer N(N<=100,000), which indicates the number of monkeys. And then N lines follows. There is one number on each line, indicating the strongness value of ith monkey(<=32768).
Second part: The first line contains an integer M(M<=100,000), which indicates there are M conflicts happened. And then M lines follows, each line of which contains two integers x and y, indicating that there is a conflict between the Xth monkey and Yth.
First part: The first line contains an integer N(N<=100,000), which indicates the number of monkeys. And then N lines follows. There is one number on each line, indicating the strongness value of ith monkey(<=32768).
Second part: The first line contains an integer M(M<=100,000), which indicates there are M conflicts happened. And then M lines follows, each line of which contains two integers x and y, indicating that there is a conflict between the Xth monkey and Yth.
Output
For each of the conflict, output -1 if the two monkeys
know each other, otherwise output the strongness value of the strongest monkey
in all friends of them after the duel.
Sample Input
5
20
16
10
10
4
5
2 3
3 4
3 5
4 5
1 5
Sample Output
8 5 5 -1 10
分析
左偏树，对于每次操作，取出最大的元素，将它从删除（即合并他的两个子树），除2后在加入进去。然后操作完成后，将两棵树合并即可。
code
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 5 using namespace std; 6 7 const int N = 100100; 8 9 int val[N],dis[N],ls[N],rs[N],fa[N]; 10 11 inline void read(int &x) { 12 x = 0;int f = 1;char ch = getchar(); 13 for (; ch<'0'||ch>'9'; ch = getchar()) if (ch=='-') f = -1; 14 for (; ch>='0'&&ch<='9'; ch = getchar()) x = x * 10 + ch - '0'; 15 x = x * f; 16 } 17 int merge(int x,int y) { 18 if (!x || !y) return x + y; 19 if (val[x]<val[y]||(val[x]==val[y]&&x<y)) swap(x,y); 20 rs[x] = merge(rs[x],y); 21 fa[rs[x]] = x; 22 if (dis[ls[x]] < dis[rs[x]]) swap(rs[x],ls[x]); 23 if (rs[x]) dis[x] = dis[rs[x]] + 1; 24 else dis[x] = 0; 25 return x; 26 } 27 inline int find(int x) { 28 if (x==fa[x]) return x; 29 return fa[x] = find(fa[x]); 30 } 31 inline int work(int x) { 32 fa[ls[x]] = ls[x];fa[rs[x]] = rs[x]; //- 33 int t = merge(ls[x],rs[x]); 34 ls[x] = rs[x] = 0; //- 35 val[x] /= 2; 36 return merge(x,t); 37 } 38 int main() { 39 int m,n,a,b; 40 while (~scanf("%d",&n)) { 41 memset(rs,0,sizeof(rs)); 42 memset(ls,0,sizeof(ls)); 43 memset(dis,0,sizeof(dis)); 44 for (int i=1; i<=n; ++i) read(val[i]),fa[i] = i; 45 read(m); 46 while (m--) { 47 read(a),read(b); 48 int x = find(a),y = find(b); 49 if (x == y) puts("-1"); 50 else printf("%d\n",val[merge(work(x),work(y))]); 51 } 52 } 53 return 0; 54 }
