time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
The ICM ACPC World Finals is coming! Unfortunately, the organizers of the competition were so busy preparing tasks that totally missed an important technical point — the organization of electricity supplement for all the participants workstations.
There are n computers for participants, the i-th of which has power equal to positive integer pi. At the same time there are m sockets available, the j-th of which has power euqal to positive integer sj. It is possible to connect the i-th computer to the j-th socket if and only if their powers are the same: pi = sj. It is allowed to connect no more than one computer to one socket. Thus, if the powers of all computers and sockets are distinct, then no computer can be connected to any of the sockets.
In order to fix the situation professor Puch Williams urgently ordered a wagon of adapters — power splitters. Each adapter has one plug and one socket with a voltage divider between them. After plugging an adapter to a socket with power x, the power on the adapter’s socket becomes equal to , it means that it is equal to the socket’s power divided by two with rounding up, for example and .
Each adapter can be used only once. It is possible to connect several adapters in a chain plugging the first to a socket. For example, if two adapters are plugged one after enother to a socket with power 10, it becomes possible to connect one computer with power 3 to this socket.
The organizers should install adapters so that it will be possible to supply with electricity the maximum number of computers c at the same time. If there are several possible connection configurations, they want to find the one that uses the minimum number of adapters u to connect c computers.
Help organizers calculate the maximum number of connected computers c and the minimum number of adapters u needed for this.
The wagon of adapters contains enough of them to do the task. It is guaranteed that it’s possible to connect at least one computer.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 200 000) — the number of computers and the number of sockets.
The second line contains n integers p1, p2, …, pn (1 ≤ pi ≤ 109) — the powers of the computers.
The third line contains m integers s1, s2, …, sm (1 ≤ si ≤ 109) — the power of the sockets.
Output
In the first line print two numbers c and u — the maximum number of computers which can at the same time be connected to electricity and the minimum number of adapters needed to connect c computers.
In the second line print m integers a1, a2, …, am (0 ≤ ai ≤ 109), where ai equals the number of adapters orginizers need to plug into the i-th socket. The sum of all ai should be equal to u.
In third line print n integers b1, b2, …, bn (0 ≤ bi ≤ m), where the bj-th equals the number of the socket which the j-th computer should be connected to. bj = 0 means that the j-th computer should not be connected to any socket. All bj that are different from 0 should be distinct. The power of the j-th computer should be equal to the power of the socket bj after plugging in abj adapters. The number of non-zero bj should be equal to c.
If there are multiple answers, print any of them.
Examples
input
2 2
1 1
2 2
output
2 2
1 1
1 2
input
2 1
2 100
99
output
1 6
6
1 0
【题解】
将插口按照power值升序排;
顺序处理每个power,如果有和当前处理的插头power相同的电脑。那么就把那个电脑插到这个插头上。如果没有，就一直除2，直到可以或power=0。为什么可以遇到就插上去呢?如果插在后面的插头，虽然可能也可以，但是必然要多耗费除2的次数(升序排的意义在此);而要满足除2次数最少则必然是这样的;
心里有些疑问吧
/*比如某个插头
power为8
同时有两个电脑power分别为4和1；
那我们肯定是把power为4的电脑插在这个插头上的(只除一次2);
而那个power为1的电脑，如果后面还有插头;
比如power=16;- >除4次;
总共5次
那是不是一开始power为8的的时候先插这个power为1的好点?
1 -> 8 :3次
后面4的时候
4 ->16:2次
总共5次
结果一样;
再试下
插头为32 64
电脑为2和16
先插16的
32
16 ->32 ->1
64
2->64 ->5
先插2的
32
2 -> 32 -> 4
64
16->64 >2
最后结果也是一样的;
但是如果没有后面那个插头显然先处理高power的更优;因为减少的次数少;
用map来快速判断有没有某个电脑的power的为x
*/#include <cstdio>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;
const int MAXN = 300000;
struct abc
{
int x, pos;
};
map <int, int> fre;
vector <int> v[MAXN];
int n, m, ans1[MAXN] = { 0 }, ans2[MAXN] = { 0 }, cnt = 0;
abc a[MAXN];
bool cmp(abc a, abc b)
{
return a.x < b.x;
}
int main()
{
//freopen("F:\rush.txt", "r", stdin);
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
{
int x;
scanf("%d", &x);
int temp = fre[x];
if (temp == 0)
{
fre[x] = ++cnt;
v[cnt].push_back(i);
}
else
v[temp].push_back(i);
}
for (int i = 1; i <= m; i++)
scanf("%d", &a[i].x), a[i].pos = i;
sort(a + 1, a + 1 + m,cmp);
int total = 0, num = 0;
for (int i = 1; i <= m; i++)
{
int sub = 0;
int x = a[i].x;
while (x)
{
int xx = fre[x];
if (xx)
{
int len = v[xx].size();
if (len)
{
int it = v[xx][len - 1];
ans2[it] = a[i].pos;
ans1[a[i].pos] = sub;
v[xx].pop_back();
total += sub;
num++;
break;
}
else
fre.erase(x);
}
if (x == 1)
break;
x = (x + 1) >> 1;
sub++;
}
}
printf("%d %d
", num, total);
for (int i = 1; i <= m; i++)
printf("%d%c", ans1[i], i == m ? '
' : ' ');
for (int i = 1; i <= n; i++)
printf("%d%c", ans2[i], i == n ? '
' : ' ');
return 0;
}