HDU 1512 HDOJ 1512 Monkey King ( 左偏�?) ACM 1512 IN HDU

MiYu — Sun, 24 Oct 2010 03:43:00 GMT

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题目地址:

http://acm.hdu.edu.cn/showproblem.php?pid=1512

题目描述 :

代码

Monkey King

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 914 Accepted Submission(s): 426

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.

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.

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

题目分析:

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Author By : MiYu

Test : 1

Complier : g++ mingw32-3.4.2

Program : HDU_1512

Doc Name : Monkey King

题目意�?

有N只猴�? 每只都有一个力量�? 开始的时候互不认�? 它们之间会发生M�ơ斗�? 每次发生a, b的斗争时, a, b都会从各自的朋友圈里拉出一个最强的, 之后两只猴子�? 打完后这两只猴子的力量值各减半. �q�且打完�? 两只猴子的朋友圈的所有�h都互相认�?也就是不会再�?.

你的��d��是对于每个斗争, 若a, b是朋�? 那么输出-1, 否则输出打完后它们的朋友圈的最强猴子的力量�?

使用普�?优先队列的话估计会超�? 因�ؓ数据量很�?100000 ! !, �{�下有空试试�?

对于每一个节�? 定义dis 表示X节点到最双��的空节点的距��ȝ��最��?/div>

对于每个节点X, 要求X的左儿子的dis >= 叛_��子的dis, 那么�Ҏ��发现, 对于N个节点的左偏�? 其右儿子最多只有logN个节�?

合�ƈ操作��是让复杂度落在叛_��子上, 从而达到logN的合�q�复杂度.

首先对于两个�? 若其中一个�ؓ�I? �q�回另一�?

否则(�q�里以大根堆��Z��), a指向堆顶较大的堆, b指向另一�? 让a的右儿子和b合�ƈ, 合�ƈ后的子树作�ؓa的右儿子.

接下�? ��查a的两个儿子是否满��dis, 不满��_��交换两个儿子.

最�? 更新a的dis.

�q�样��容易实现堆的其他操�?( 比如插入, 删除��等 ).

另外 �q�需要用�?�q�查�?

//#pragma warning( disable:4789 )

#include

using namespace std;

const int MM = 100010;

struct left {

int l,r,dis,val,dad;

} heap[MM];

int N, M;

inline int max ( const int &a, const int &b) {

return a > b ? a : b;

}

inline int find ( int &x ) {

return heap[x].dad == x ? x : heap[x].dad = find ( heap[x].dad );

}

inline void swap(int &a, int &b) {

a ^= b ^= a ^= b;

}

inline int merge ( int x, int y ) {

if ( x == 0 ) return y;

if ( y == 0 ) return x;

if ( heap[y].val > heap[x].val ) swap ( x, y );

heap[x].r = merge ( heap[x].r, y );

heap[heap[x].r].dad = x;

if ( heap[ heap[x].l ].dis < heap[ heap[x].r ].dis )

swap ( heap[x].l, heap[x].r );

if ( heap[x].r == 0 ) heap[x].dis = 0;

else heap[x].dis = heap[ heap[x].r ].dis + 1;

return x;

}

inline int push ( int x, int y ) {

return merge ( x, y );

}

inline int pop ( int &x ) {

int l = heap[x].l;

int r = heap[x].r;

heap[l].dad = l;

heap[r].dad = r;

heap[x].l = heap[x].r = heap[x].dis = 0;

return merge ( l, r );

}

inline bool scan_d(int &num) {

char in;bool IsN=false;

in=getchar();

if(in==EOF) return false;

while(in!='-'&&(in<'0'||in>'9')) in=getchar();

if(in=='-'){ IsN=true;num=0;}

else num=in-'0';

while(in=getchar(),in>='0'&&in<='9'){

num*=10,num+=in-'0';

}

if(IsN) num=-num;

return true;

}

int main() {

while ( scan_d ( N ) ) {

for ( int i = 1; i <= N; ++ i ) {

scan_d ( heap[i].val );

heap[i].l = heap[i].r = heap[i].dis = 0;

heap[i].dad = i;

}

scan_d ( M );

int a, b, x, y;

while ( M -- ) {

scan_d (a); scan_d (b);

x = find ( a );

y = find ( b );

if ( x == y ) {

puts ( "-1" );

} else {

heap[x].val /= 2;

int xx = push ( pop ( x ), x );

heap[y].val /= 2;

int yy = push ( pop ( y ), y );

printf ( "%d\n", heap[ merge ( xx, yy ) ].val );

}

return 0;

}

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HDU 1512 HDOJ 1512 Monkey King ( 左偏�?) ACM 1512 IN HDU