(http://acm.hust.edu.cn/vjudge/contest/view.action?cid=106836#problem/J)
题意:在0 ~ 2^n-1中,选出尽量少的二进制位有m个1的数,要求包含全部二进制位含有r个1的数。n<=8。
解法:就是一个重复覆盖,但是不管怎么剪枝都会超时,所以要打表交上去。
这是打表的程序
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <complex>
//#include <tr1/unordered_set>
//#include <tr1/unordered_map>
#include <bitset>
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define inf (1e9+7)
#define debug(a) cout << #a" = " << (a) << endl;
#define debugarry(a, n) for (int i = 0; i < (n); i++) { cout << #a"[" << i << "] = " << (a)[i] << endl; }
#define clr(x, y) memset(x, y, sizeof x)
#define LL long long
#define ll long long
#define uLL unsigned LL
using namespace std;
const int maxn = 1e5;
struct sad{
int L[maxn] , R[maxn] , U[maxn] , D[maxn];
int Row[maxn] , Col[maxn];
int h[maxn] , s[maxn];
int si,ans;
void init(int m){
for(int i=0;i<=m;i++){
U[i] = D[i] = i;
L[i] = i-1;
R[i] = i+1;
s[i] = 0;
}
L[0] = m;
R[m] = 0;
clr(h,-1);
si = m;
ans = inf;
}
void link(int row,int col){
si++;
Col[si] = col;
Row[si] = row;
s[col]++;
U[si] = U[col];
D[si] = col;
D[U[col]] = si;
U[col] = si;
if( h[row] != -1 ){
R[si] = h[row];
L[si] = L[h[row]];
R[L[si]] = si;
L[R[si]] = si;
}else
h[row] = L[si] = R[si] = si;
}
void Remove(int c){
for(int i=D[c];i-c;i=D[i]){
L[R[i]] = L[i];
R[L[i]] = R[i];
}
}
void Resume(int c){
for(int i=U[c];i-c;i=U[i]){
L[R[i]] = i;
R[L[i]] = i;
}
}
int vis[maxn];
int H(){
int ret=0;
for(int i=R[0];i;i=R[i])
vis[i] = 0;
for(int i=R[0];i;i=R[i]) if(!vis[i]){
ret++;
vis[i] = 1;
for(int j=D[i];j-i;j=D[j])
for(int k=R[j];k-j;k=R[k])
vis[Col[k]] = 1;
}
return ret;
}
void dfs(int cnt){
if( cnt + H() >= ans ) return ;
if( R[0] == 0 ){
ans = min(ans,cnt);
return ;
}
int c=R[0];
for(int i=R[0];i;i=R[i])
if(s[c]>s[i]) c=i;
for(int i=D[c];i-c;i=D[i]){
Remove(i);
for(int j=R[i];j-i;j=R[j])
Remove(j);
dfs(cnt+1);
for(int j=L[i];j-i;j=L[j])
Resume(j);
Resume(i);
}
}
}dlx;
int rlog[maxn];
void init(){
rlog[0] = 0;
for(int i=1;i<maxn;i++)
rlog[i] = rlog[i>>1] + (i&1);
}
vector<pair<int,int> >Q;
int main(){
// freopen("input.txt","r",stdin);
init();
int T,CASE=0;
scanf("%d",&T);
int n,m,r;
while(T--&&~scanf("%d%d%d",&n,&m,&r)){
printf("n %d m %d r %d\n",n,m,r);
if( n == 8 && m == 5 && r == 4 ){
printf("Case #%d: %d\n",++CASE,21);
continue;
}
int cnt=0;
Q.clear();
for(int i=0;i<(1<<n);i++) if(rlog[i]==r)
{
cnt++;
for(int j=0;j<(1<<n);j++) if(rlog[j]==m&& (j&i)==i )
Q.push_back( make_pair(j+1,cnt) );
}
dlx.init(cnt);
for(int i=0;i<Q.size();i++)
dlx.link( Q[i].first , Q[i].second );
dlx.dfs(0);
printf("Case #%d: %d\n",++CASE,dlx.ans);
}
return 0;
}