2018 ACM-ICPC, Syrian Collegiate Programming Contest F. Pretests(子集dp)

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题意

一道题有kkk个测试点,共提交了nnn次

每次提交是一个长为kkk的010101串,表示是否能通过第iii个测试点

现在要求你重新排序kkk个测试点,使得∑i=1nw[i]\sum\limits_{i=1}^nw[i]i=1∑nw[i]最小

其中w[i]w[i]w[i]指的是第iii个人第一次在第几个测试点wronganswerwrong\ answerwrong answer了

输出字典序最小的方案数

考虑定义f[i]f[i]f[i]表示当放置测试点状态为iii时,能得到的最小代价

那么枚举本次放的是第jjj道题,放置之后由状态vvv转移到状态uuu

造成的代价是num(u)num(u)num(u)(uuu的二进制位是111的个数,表示在这个测试点wawawa掉)

因为在之前的num(u)−1num(u)-1num(u)−1个测试点都没有wawawa掉,在第num(u)num(u)num(u)个测试点wawawa掉

我们需要求满足要求的提交个数也就是uuu的超集,再减去vvv的超集,那么计算出这次转移的代价valvalval

f[u]=min(f[u],f[v]+val)f[u]=min(f[u],f[v]+val)f[u]=min(f[u],f[v]+val)

记录一个前驱就可以转移了

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e6+10;
const int inf = 1e9;
int t,n,T,num[maxn],pre[maxn],cnt[maxn],f[maxn],bit[maxn];
string s[maxn];
void sosdp()
{for(int i=1;i<=n;i++)   cnt[num[i]]++;for(int i=0;i<t;i++)for(int j=0;j<(1<<t);j++)if( !(j&(1<<i)) )  cnt[j] += cnt[j|(1<<i)];
}
void dfs(int sta)
{if( sta==(1<<t)-1 )    return;printf("%d ",pre[sta] );dfs( sta | (1<<(pre[sta]-1)) );
}
void DP()
{sosdp();for(int i=0;i<(1<<t);i++)  f[i] = inf;f[(1<<t)-1] = 0;for(int i=(1<<t)-1;i>=0;i--){for(int j=1;j<=t;j++)//枚举每一个测试点{if( i&(1<<(j-1)) ){int v = i ^ (1<<(j-1)), val = f[i] + ( cnt[v]-cnt[i] )*bit[i];if( val<f[v] ) f[v] = val, pre[v] = j;else if( val==f[v] ) pre[v] = min( pre[v],j );}}    }for(int i=1;i<=n;i++)   if( num[i]==(1<<t)-1 )  f[0] += t;cout << f[0] << endl;dfs( 0 );  cout << endl;
}
void init()
{for(int i=0;i<(1<<t);i++)  cnt[i] = 0;
}
int main()
{freopen("tests.in","r",stdin);bit[1] = 1;for(int i=2;i<=2000000;i++)   bit[i] = bit[i>>1] + (i&1);cin >> T;while( T-- ){cin >> t >> n;for(int i=1;i<=n;i++){cin >> s[i];int x = 0;for(int j=0;j<t;j++)   if( s[i][j]=='1' )  x |= (1<<j);num[i] = x;}DP();init();}
}

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